Question

When 66.8g of (O2) gas is mixed with 27.8g of (NH3) gas and 25.1g of (CH4) gas, 36.4g of (HCN) gas is produced. What is the percent yield of HCN in this reaction?

Answer 1

Well, we need a stoichiometric equation...

Explanation:

`2CH_4(g) + 2NH_3(g) + 3O_2(g) rarr2HC-=N(g) + 6H_2O(l)`

For the life of me I cannot remember the name of this reaction, but this stoichometric equation really should have been supplied with the question....

`"Moles of hydrogen cyanide"=(36.4*g)/(27.03*g*mol^-1)=1.35*mol`

`"Moles of dioxygen"=(66.8*g)/(32.00*g*mol^-1)=2.09*mol`

`"Moles of ammonia"=(27.8*g)/(17.03*g*mol^-1)=1.63*mol`

`"Moles of methane"=(25.1*g)/(16.01*g*mol^-1)=1.57*mol`

Anyway, I leave it to you to calculate the percentage yield based on the stoichiometry....