Question

What is the percent yield of a reaction in which 762g CH4 id converted to 2048g CH3Cl? The equation for the reaction is CH4+Cl2=CH3Cl+HCl

Answer 1

I'm assuming this is a limiting reagent question if you're asking for the percent yield of CH3Cl . Is that all the information given? Is there a value for Cl2?

Answer 2

only
multiple choice answers: 2.69% 26.9% 37.2% 85.4% 117%
Answer is 85.4%, but don't know how to get it. Thanks

Answer 3

It's quite easy we found that 2048g of CH3Cl was actually produced during the reaction. Now you just need to find how much CH3Cl would be produced if the reaction went to 100% completion. We use the mass of CH4 to find how many moles \[762g \div 16.04g/mol\] to find that 47.51 mols of CH4 are produced, using the ratio in our complete reaction we get that \[ CH _{4} = CH _{3}Cl \] in a 1:1 mole ratio. So there would be 47.51 mol CH3Cl is equation went to 100% completion . Then we convert the moles to mass . \[47.51 mol CH _{3}Cl \times 50.49g/mol\] where 50.49 is the molar mass. Give us 2348.5grams. Thats our theoretical yield if it went to completion .
Then we just use our equation for percent yield \[percent yield = (actual mass/ theoretical mass ) \times 100\] so \[(2048g \div 2398.5 )\times 100\] giving you 85.3867 = 85.4%