chem question?

Question

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane:

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

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

Answer 1

5,000,000g NH3 / 17g/mol = 294,117.6471molNH3
5,000,000gO2 / 32g/mol = 156,250molO2
5,000,000gCH4 / 16g/mol = 312,500molCH4
2/294,117.6471 = 2/x, x = 294,117.6471molHCN produced
3/156,250 = 2/x, x = 104,166.6667molHCN produced
2/312,500 = 2/x, x = 312,500molHCN produced
O2 is limiting reagent.
3/156,250 = 6/x, x = 312,500molH2O produced
104,166.6667molHCN * 27g/mol = 2.812500001 x 10^6 g HCN produced
312,500molH2O * 18g/mol = 5.625 x 10^6 g HCN produced

Answer: 2.812500001 x 10^6 g H2O
5.625 x 10^6 g H2O

Answer 2

Determine your limiting reactant by calculating the number of moles of each reactant.

5.00 * 10^6 g / 17 g/mole NH3 = 2.94 * 10^5 moles NH3

5.00 * 10^6 g / 32 g/mole O2 = 1.56 * 10^5 moles O2

5.00 * 10^6 g/ 16 g/mole CH4 = 3.13 * 10^5 moles CH4

Since you need 1.5 times as many moles of O2 as you do NH3 or CH4 by the equation and you don't have it, O2 is your limiting reactant. You will generate 2/3 as many moles of HCN and twice as many moles of H2O as you had O2 to start with.

1.56 *10^5 moles O2 * (2 moles HCN/3 moles O2) * 27 g/mole = 2.81 *10^6 g = 2.81 *10^3 kg HCN

1.56 * 10^5 moles O2 * (6 moles H2O/3 moles O2) * 18 g/mole = 5.62 *10^6 g = 5.62 * 10^3 kg H2O