If 5x10^3 kg each of NH3,O2 & CH4 are reacted,what mass of HCN & OF H2O will be produced, assuming 100% yield?

Question

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxegen, and methain:

2NH3(g) + 3O2(g) + 2CH4(g)-----> 2HCN(g) + 6H2O(g)

If 5.00x 10^3 kg each of NH3, O2 and of CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield??

HELP!!!!!!

Answer 1

First we compute the number of moles of each substance to determine the limiting reagent.

First we convert 5.00x10^3 kg into grams:

5.00x10^3 kg * 1000g/1kg = 5.00x10^6 g

NH3 has a molecular mass of 17.01 g/mol

Now we convert to moles:

(17.01 g/mol)^-1 = 5.879 x 10-2 mol/g * 5.00x10^6 g

= 2.94 x 10^5 mols NH3

O2 has molecular mass 32.00 g/mol. So a similiar calculation as the one above yields:

1.56 x10^5 mol O2

For CH4 we get mass 16.00 g/mol and 3.13 x10^5 mol CH4

In this case the limiting reagent is O2 since all of the 1.56 x10^5 mols of O2 will be consumed while only (1.56x10^5)(3/2) = 2.34x10^5 mols of the other two compounds will be used.

Having the limiting reagent we can now calculate the amount of H2O with 100% yield.

1.56x10^5 mol O2 means we will have 2*(1.56x10^5) mol H20

H20 has a molecular mass of 18.00 g/mol

18.00 g/mol  *  2*(1.56x10^5) mol = 5.62x10^6 g H20

Looking at the right side of the original equation we have 2 moles of HCN for every 6 of H20.

Thus we must get get (1/3)*(2)*(1.56x10^5) mol HCN

That is 1.04x10^5 mol HCN with a molecular mass of 27 g/mol.

(27 g/mol)*(1.04x10^5 mol) = 2.8 x 10^6 g HCN

2.8x10^3 kg HCN  5.62x10^3 kg H20