Question

Methanol, CH3 OH, is made by reacting carbon monoxide and hydrogen gas. It is used

for fuel in race cars. If 8.60 kg of hydrogen gas are reacted with excess carbon monoxide, how much methanol should be produced? If the actual yield is 3.57 x 10^4 grams, what is the percent yield?

Answer 1

CO + 2H2 = CH3OH

n(H2) = m/M = 8600g/2g/mol = 4300 mol

n(CH3OH) = 4300 mol

m(CH3OH) = n*M = 4300mol*32g/mol = 137600 g = 13.76*10^4 g

Percent yield = 3.57*10^4/13.76*10^4 = 0.259 or 25.9%