How do you use stoichiometry to convert moles to mass?

1 Answer
Aug 4, 2018

Answer:

Suppose we completely combusted an #8*g# mass of methane...

Explanation:

Clearly, we could write the stoichiometric reaction, the which gives us the absolute molar equivalence with respect to product and reactants...

#underbrace(CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(l) +Delta)_"as written 80 g of reactant gives 80 g of product"#

But we proposed an #8*g# mass with respect to methane, a molar quantity of #(8*g)/(16.01*g*mol^-1)=1/2*mol#...

Given the stoichiometry, therefore we require a #1*mol# quantity of dioxygen gas, i.e. #32*g#, with which the methane reacts to give a #1/2*mol# quantity #CO_2#, i.e. #22*g#, and ONE mole of water, i.e. #18*g#...and thus #40*g# of reactant gives PRECISELY #40*g# of product....PLUS ENERGY.