Previously, we explored the Stoichiometry of hydrogen powered bicycles that "run on water". They typically have a rechargeable battery pack and electric hub motor. The company that makes the electrical fuel cell won a Presidential Green Chemistry Award [1] . Electric hub motors on front or back wheels Now we'll see how the water cannister and fuel canister need to be matched for maximum range. In the laboratory as well as the environment, inexpensive reagents like atmospheric O 2 or water are often supplied in excess. Some portion of such a reagent will be left unchanged after the reaction. Conversely, at least one reagent is usually completely consumed. When it is gone, the other excess reactants have nothing to react with and they cannot be converted to products. The substance which is used up first is the limiting reagent . A new electricity source combines a hydrogen fuel cell with a "sodium silicide" fuel cartridge, winner of a "Green Chemistry Challenge Award" [2] . The sodium silicyde reacts with water to make the hydrogen fuel [3] [4] [5] : 2 NaSi(s) + 5H 2 O(l) → Na 2 Si 2 O 5 (s) + 5H 2 (g) (1) The composition of sodium silicide may depend on the method of synthesis. Silicides can be made by the reaction of active metals (like Mg) with sand, or by heating sodium with silicon. Dye et al [6] prepare sodium silicide by the reaction of sodium metal with silica gel, obtaining black powders of (hypothetically) Na 4 Si 4 nanoparticles. 0 Contents 1 Example 1 2 General Strategy for Limiting Reactant Problems 3 Example 2 4 References Example 1 Sodium silicide/water bikes produce about 200W to run a bicycle for about 30 miles with a NaSi cartridge weighing about 1.5 lb [7] . If a bicyclist wants to travel about 30 miles and brings 1 quart of water and one 1.5 lb cartridge, which is the limiting reagent? What mass of solid product will be formed? Solution The balanced equation 2 NaSi(s) + 5H 2 O(l) → Na 2 Si 2 O 5 (s) + 5H 2 (g) (1) tells us that according to the atomic theory, 2 mol NaSi is required for each 5 moles of 5H 2 O. That is, the stoichiometric ratio S(NaSi/H 2 O) = 2 mol NaSi/ 5 mol H 2 O. Let us calculate the amount in moles of each we actually have, assuming the density of water is 1.00 g/mL: \(\text{m}_{\text{H}_{2}\text{O}}~=~\) \(\text{1 quart}~\times~\frac{\text{1 L}}{\text{1.05668821 quart}}~\times~\frac{\text{1000 g}}{\text{1 L} }~ =~\text{1057 g}\) \(\text{m}_{\text{NaSi}}~=~ <math>\text{1.5 lb}~\times~\frac{\text{453.59 g}}{\text{ quart}}~ =~\text{680 g}\) \(\text{n}_{\text{NaSi}}=\text{680 g}\times \frac{\text{1 mol NaSi}}{\text{51.075 g}}=\text{13.3 mol NaSi}\) \(\text{n}_{\text{H}_{\text{2}}\text{O}}=\text{1057 g}\times \frac{\text{1 mol H}_{\text{2}}\text{O}}{\text{18.015 g}}=\text{58.57 mol H}_{\text{2}\text{O}}\) If all the H 2 O were to react, the stoichiometric ratio allows us to calculate the amount of NaSi that would be required: \(\text{n}_{\text{NaSi}}~=~ \text{n}_{\text{H}_{2}\text{O}}~~\times~~\) \(~~\frac{\text{2 mol NaSi}}{\text{5 mol H}_{2}\text{O}}~~\) \(~=~\text{23.5 mol NaSi}\) This is more than the amount of NaSi present,so NaSi is the limiting reactant and H 2 O is present in excess. If all the NaSi reacts, the stoichiometric ratio allows us to calculate the amount of H 2 O that would be required: \(\text{n}_{\text{H}_{2}\text{O}}~=~ \text{n}_{\text{NaSi}}~~\times~~\) \(~~\frac{\text{5 mol H}_{2}\text{O}}{\text{2 mol NaSi}}~~\) \(~=~\text{13.3 mol NaSi}\) \(~\times~\frac{\text{5 mol H}_{2}\text{O}}{\text{2 mol NaSi}}~=~\) 33.3 mol H 2 O We require less than the amount of H 2 O present, so it is the excess reactant . When the reaction ends, 13.3 mol of NaSi will have reacted with 33.3 mol H 2 O and there will be (58.57 mol - 33.3 mol) = 25.4 mol H 2 O left over. NaSi is therefore the limiting reagent. b. Since the water doesn't all react, we need to calculate the amount of solid product produced from the amount of NaSi consumed, by using the stoichiometric ratio: \(\text{n}_{\text{Na}_{2}\text{Si}_{2}\text{O}_{5}}~=~ \text{n}_{\text{NaSi}}~~\times~~\) \(~~\frac{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{5}}{\text{2 mol NaSi}}~~\) \(~=~\text{6.65 mol Na}_{2}\text{Si}_{2}\text{O}_{5}\) The mass of water is then calculated by using the molar mass: \(\text{6.65 mol Na}_{2}\text{Si}_{2}\text{O}_{5}~\times~\) \(\frac{\text{182.148 g}}{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{5}}\) \(~=~\text{1211 g Na}_{2}\text{Si}_{2}\text{O}_{5}\) c. About a pound (458 g) of water will react, and 1057-458 = 599 g of water will remain. That's enough for another 1.5 lb canister of NaSi. These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting. We use the amount of limiting reagent to calculate the amount of product formed. Unnamed: 0 2 NaSi + 5 H2O → 1 Na2Si2O5 + 5 H2 m (g) 680.000 1057.000 NaN NaN M (g/mol) 51.075 18.015 182.148 2.016 n (present,mol) 13.300 58.570 NaN NaN if all H2O reacts -58.670 -23.500 11.730 NaN if all NaSi reacts -13.300 -33.300 6.650 33.300 Actual Reaction Amounts -13.300 -33.300 6.650 33.300 Actual Reaction Masses -680.000 -600.000 1211.000 67.200 In the end, 1057 - 599 = 458 g of H 2 O will remain, along with 1211 g of sodium silicate. The 67.2 g of hydrogen produced gives a total of 1736 g. The mass of the reactants was also 1057 + 680 = 1737 g, equal within the error of measurement. General Strategy for Limiting Reactant Problems From this example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting. We must compare the stoichiometric ratio S(X/Y) with the actual ratio of amounts of X and Y which were initially mixed together. The general rule, for any reagents X and Y, is \(\begin{align} & \text{If}~ \frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is less than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then X is limiting}\text{.} \\ & \\ & \text{If}~\frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is greater than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then Y is limiting}\text{.} \\ \end{align}\) (Of course, when the amounts of X and Y are in exactly the stoichiometric ratio, both reagents will be completely consumed at the same time, and neither is in excess.). This general rule for determining the limiting reagent is applied in the next example. Example 2 Suppose the sodium silicide made by a different method is Na 2 Si [8] [9] , and it produces ordinary "water glass", or sodium silicate (Na 2 SiO 3 (s) [10] ) rather than Na 2 Si 2 O 5 (s) [11] . In that case, if the bicyclist wants to travel about 30 miles and brings 1 quart of water and one 1.5 lb cartridge, which is the limiting reagent? Solution The balanced equation is: Na 2 Si(s) + 3H 2 O(l) → Na 2 SiO 3 (s) + 3H 2 (g) (1) Now the stoichiometric ratio S(NaSi/H 2 O) = 1 mol Na 2 Si/ 3 mol H 2 O. Again, let's calculate the amount in moles of each we actually have, assuming the density of water is 1.00 g/mL: \(\text{m}_{\text{H}_{2}\text{O}}~=~\) \(\text{1 quart}~\times~\frac{\text{1 L}}{\text{1.05668821 quart}}~\times~\frac{\text{1000 g}}{\text{1 L} }~ =~\text{1057 g}\) \(\text{m}_{\text{NaSi}}~=~ <math>\text{1.5 lb}~\times~\frac{\text{453.59 g}}{\text{ quart}}~ =~\text{680 g}\) \(\text{n}_{\text{Na}_{2}\text{Si}}~=~\) \(\text{680 g}\times \frac{\text{1 mol Na}_{2}\text{Si}}{\text{74.065 g}}~=~\) 9.18 mol Na 2 Si \(\text{n}_{\text{H}_{\text{2}}\text{O}}=\text{1057 g}\times \frac{\text{1 mol H}_{\text{2}}\text{O}}{\text{18.015 g}}=\text{58.57 mol H}_{2}\text{O}\) If all the H 2 O were to react, the stoichiometric ratio allows us to calculate the amount of NaSi that would be required: \(\text{n}_{\text{Na}_{2}\text{Si}}~=~\) \(\text{n}_{\text{H}_{2}\text{O}}~~\times~~\) \(~~\frac{\text{1 mol Na}_{2}\text{Si}}{\text{3 mol H}_{2}\text{O}}~~\) \(~=~\text{19.52 mol Na}_{2}\text{Si}\) This is more than the amount of Na 2 Si present,so Na 2 Si is the limiting reactant and H 2 O is present in excess. If all the Na 2 Si reacts, the stoichiometric ratio allows us to calculate the amount of H 2 O that would be required: \(\text{n}_{\text{H}_{2}\text{O}}~=~\) \(\text{n}_{\text{Na}_{2}\text{Si}}~~\times~~\) \(~~\frac{\text{3 mol H}_{2}\text{O}}{\text{1 mol Na}_{2}\text{Si}}~~\) \(~=~\text{13.3 mol Na}_{2}\text{Si}\) \(~\times~\frac{\text{3 mol H}_{2}\text{O}}{\text{1 mol Na}_{2}\text{Si}}~=~\) 27.54 mol H 2 O We require less than the amount of H 2 O present, so it is the excess reactant . When the reaction ends, 9.18 mol of Na 2 Si will have reacted with 27.54 mol H 2 O and there will be: (58.57 mol - 27.54 mol) = 31.03 mol H 2 O left over. Na 2 Si is therefore the limiting reagent. These calculations can again be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting. We use the amount of limiting reagent to calculate the amount of product formed. Unnamed: 0 1 Na2Si + 3 H2O → 1 Na2SiO3 + 3 H2 m (g) 680.000 1057.000 NaN NaN M (g/mol) 74.065 18.015 122.06 2.016 n (present,mol) 9.180 58.570 NaN NaN if all H2O reacts -58.670 -19.520 58.67 19.520 if all NaSi reacts -9.180 -27.540 27.54 9.180 Actual Reaction Amounts -9.180 -27.540 27.54 9.180 Actual Reaction Masses -680.000 -496.000 1121.00 55.500 In the end, 1057 - 496 = 561 g of H 2 O will remain, along with 1121 g of sodium silicate. The 55.5 g of hydrogen produced gives a total of 1176 g. The mass of the reactants was also 496 + 680 = 1176 g, equal within the error of measurement. This process seems less efficient than the one in Example 1, because only 55.5 g of hydrogen is produced, compared to 67.2 g in Ex. 1. Also, 496 g of water and 680 g of silicide are consumed (totalling 1176 g), while in Ex. 1, 599 g water and 680 g of silicide (1280 g total) were consumed, so we get 0.047 g hydrogen/g reactants in Ex. 2, while we get 0.053 g hydrogen/g reactants in Ex. 1. References signachem.com/wp-content/them...Green_Chem.pdf signachem.com/wp-content/them...Green_Chem.pdf http://www.greencarcongress.com/2006/07/hydrogen_produc.html http://www.physorg.com/news205599186.html www.wired.com/gadgetlab/2010/...lmost-on-water James L. Dye, Kevin D. Cram, Stephanie A. Urbin, Mikhail Y. Redko, James E. Jackson, and Michael Lefenfeld J. Am. Chem. Soc., 2005, 127 (26), pp 9338–9339 Publication Date (Web): June 14, 2005 (Communication) DOI: 10.1021/ja051786+ www.wired.com/gadgetlab/2010/...most-on-water/ Greenwood, Norman N.; Earnshaw, A. (1997), Chemistry of the Elements (2nd ed.), Oxford: Butterworth-Heinemann, ISBN 0080379419 http://en.Wikipedia.org/wiki/Silicide en.Wikipedia.org/wiki/Sodium_silicate http://pubs.acs.org/doi/abs/10.1021/ie0303909