I haven't done these in some time, so I'm not sure if they are 100% right.
1) 6.3 moles of H2( 2 mol of NH3 / 3 mol of H2)= 4.2 mol of NH3
6.3 moles of H2( 2 mol of NH3/ 3 mol of H2)(17.04 g of NH3/1 mol NH3)= 71.57 g of NH3
2) 2.5 moles of N2(2 mol of NH3/1 mol of N2)= 5 moles of NH3
2.25 moles of N2(2 mol of NH3/ 1 mol of N2)(6.02x10^23 particles/ 1 mol of NH3)= 3.01x10^24 particles of NH3
3) 425 g of NH3(1 mol of NH3/17.04 g NH3)= 24.9 moles of NH3
425 g of NH3(1 mol of NH3/17.04 g of NH3)(1 mol of N2/2 mol of NH3)(28.02 g of N2/1 mol N2)= 349 g of N2
425 g of NH3(1 mol of NH3/17.04 g of NH3)(1 mol of N2/2 mol of NH3)= 12.5 mol of N2
4) 10 moles NH3(3 moles of H2/2 moles of NH3)= 15 moles H2
10 moles NH3(3 mol of H2/2 mol of NH3)(2.02 g of H2/1 mol of H2)= 30.3 g of H2
30.3 g = .0303 liters of H2
Answer:
A scientist measures the standard enthalpy change for the following reaction to be -139.5 kj :
h2(g) + c2h4(g)c2h6(g)
based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of c2h4(g) is _____ kj/mol
Explanation:
Hydrogen ΔHof (kJ/mol) ΔGof (kJ/mol) So (J/mol K)
H2 (g)
0
0
130.7
Carbon ΔHof (kJ/mol) ΔGof (kJ/mol) So (J/mol K)
C2H6 (g)
-84.7
-32.8
229.6
Going from left to right, they are ordered by increase in atomic number
