Reactions involving trimethylaluminium Al2Me6 I need to write balanced equations describing the following reactions:


one mole of $\ce{Al2Me6}$ with two moles of water
excess of $\ce{Al2Me6}$ with silicon dioxide
excess of $\ce{Al2Me6}$ with tin(IV) chloride


My answer so far:


a) $\ce{2H2O + Al2Me6 -&gt; Al2Me4(H2O)2}$?
b) $\ce{SiO2 + Al2Me6 -&gt; SiMe4 + Al2Me2O2}$?
c) $\ce{SnCl4 + Al2Me6 -&gt; SnMe4 + Al2Me2Cl4}$?
For a): Many organometallic compounds are easily hydrolyzed to the respective hydrocarbons and metal (hydr)oxides. In this case, the products are $\ce{CH4}$ and more likely $\ce{Al2O3}$, because the formation of $\ce{Al(OH)3}$ requires more moles of water. 

Reactions b) and c) should be transmetallations where the methyl groups are transferred from aluminium to the other metal. The excess of $\ce{Al2Me6}$ helps to shift the equilibrium of the reaction to the product side. $\ce{SiMe4}$ and $\ce{SnMe4}$ would be the correct products which you have identified, but aluminium is also converted into its respective oxide/chloride.