Calculate volume of hydrogen gas evolved

**MCQ Question:**


Ammonia borane in water can release hydrogen upon hydrolysis while forming NH4BO2 as the other
product.


The volume (cm3) of hydrogen gas which will be evolved from the hydrolysis of 309 mg of
ammonia borane (NH3BH3) with a purity of 90 % w/w at 27 °C and at atmospheric pressure 1.0 bar is
approximately,


(a) 66.5 cm3


(b) 224.5 cm3


(c) 673.4 cm3


(d) 893.2 cm3


(e) none of the above


The answer from the mark scheme is **C) 673.4 cm3**


I took the reaction equation as


NH3BH3 + 2 H2O -> NH4BO2 + 3 H2


However I cannot figure out how to attempt this question. What method should I follow to get 673.4 cm3 of hydrogen gas?




NH3BH3 + 2H2 -> NH4BO3 + 3H2


309mg = 0.309g - use moles = mass/molar mass equation:


mol(NH3BH3) = 0.309/14+6(1)+10.8 = 0.309/30.8 = 0.01mol


Use stochiometric ratio 1:3 to recognize that you multiply the moles of NH3BH3 by 3 to get moles of H2, 0.01mol \* 3 = 0.03mol = mol(H2)


Prepare variables for pV = nRT (pressure in Pa \* volume in m^3 = 8.314 \* moles of gas (H2) \* temperature in kelvin) , rearranging equation to give V = nRT/p


To convert degrees C into Kelvin, add 273: 27 + 273 = 300K


To convert "bar" into Pa - one "bar" = 100000Pa


R is ideal gas constant 8.314


Substitute values into equation V = (0.03) \* (8.314) \* (300) / (100000) = 7.48 \* 10^-4 m^3


Convert m^3 into cm^3 by multiplying by 1 million 7.48 \* 10^-4 m^3 = 748.26cm^3


The 'purity' thing implies that only 90% of the calculated value is the answer - 784.26 \* 0.9 = 673.4cm^-3 which is the answer.


 1