Question

How do you use stoichiometry to convert moles to mass?

Answer 1

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.