When methanol undergoes complete combustion, the products are carbon dioxide and water. 2CH3OH (l) + 3O2(g) --> 2CO2(g) + 4H2O (g). How much water ( in grams) can you maximally produce if you burn 50g of methanol? In real experiment, you isolate 10 mL of water (d=0.98 g/cm3) , what's your yield?

Answer 1

You must always have a balanced equation and you have that.

Convert 50 g methanol to moles. moles = grams/molar mass

Using the coefficients in the balanced equation, convert moles methanol to moles H2O.

Now convert moles H2O to grams. g = moles x molar mass. This is the theoretical yield.

The above is what you need to work any simple stoichiometric problem. The next step converts to yield.
%yield = (actual yield/theoretical yield)*100 = ??

You have actual yield in the problem of 10 mL water. Use the density to convert that to grams water, then substitute into the above percent formula.