Hey all. I just needed some help verifying that this is correct. Thanks all!

Use the following equation to answer the questions below:

CH4 + NH3 + O2 > HCN + H2O

Balance the equation.

Given that we have 21 g of methane (CH4) and 12 g of ammonia (NH3), which reactant is limiting? Show your calculations and explain how you determined the limiting reactant.

What is the theoretical yield of hydrogen cyanide will be formed? Show all your work.

If the actual yield of hydrogen cyanide is 53%, then how many grams of HCN is produced?

What is the actual yield of water?

Answer:

The balanced equation will be 2CH_4+2NH_3+3O_2→2HCN+6H_2 O

Given that the ratio of reactants to product in the balanced equation is 2:2:3→2:6, the limiting reactant would be ammonia, NH_3. The ratio of 2:2:3→2:6 states that 2 moles of CH_4, 2 moles of NH_3, and 3 moles of O_2 will produce 2 moles of HCN and 6 moles of H_2 O.

Find how many moles of each reactant we have by determining the molar weight.

CH_4 moles=21/[C+(H*4)]

CH_4 moles=21/[12+(1*4)]

CH_4 moles≈1.3

NH_3 moles=12/([N+(H*3)])

NH_3 moles=12/([14+(1*3))

NH_3 moles≈0.7

The moles of ammonia and methane should be equal for there to be no limiting reactant, but as there is less moles of ammonia, it is the limiting reactant.

To find the theoretical yield of HCN, first find how many moles of the limiting reactant is present in 12g. Since the limiting reactant is NH_3, find its molar weight, in g/mol.

The weight of N is about 14 amu. The weight of H is about 1 amu. Combine: 14+(1*3)=17 g/mol.

Divide 12 by 17: 12/17≈0.7

Since the ratio of NH_3 to HCN is 2:2, or 1:1, 0.7 moles of NH_3 will produce 0.7 moles of HCN.

Find the molar weight of HCN. The mass of hydrogen (H) is about 1 amu. The mass of carbon (C) is about 12 amu. The mass of nitrogen (N) is about 14 amu. Combine: 1+12+14=27g/mol.

Multiply the molar mass by the number of moles: 0.7*27=18.9g.

Thus, there is about 18.9 grams produced with a theoretical yield of 100%.

If the actual yield of hydrogen cyanide is 53%, when provided with 2 moles of NH_3, 2*0.53 or 1.06 moles of HCN will be produced. The ratio is now 2:1.06

As we are given 0.7 moles of NH_3, multiply it by the fraction 1.06/2 to find the moles produced of HCN: 0.7*1.06/2=0.371. Thus, 0.371 moles of HCN are produced with a yield of 53%.

Multiply the moles by the molar weight of HCN: 0.371*27=10.017.

We could also find the actual yield by multiplying the theoretical yield (18.9) by 53% (0.53): 18.9*0.53=10.017.

Thus, there are about 10.017 grams of hydrogen cyanide produced with a yield of 53%

The ratio of the first product, HCN, to H_2 O is 2:6, or 1:3.

To find the actual yield of water when given the actual yield of hydrogen cyanide (10.017 grams), multiply the amount of HCN by 3.

10.017*3=30.051

Thus, there is about 30.051 grams of actual water yielded.