Answer:
The ΔHrxn for the above equation = 179 kJ/mol
Explanation:
The reaction bond enthalpies are for the reactant;
3 × N-H = 3 × 390 = 1,170 kJ/mol
2 × O=O = 2 × 502 = 1004 kJ/mol
The reaction bond enthalpies are for the product;
3 × N-O = 3 × 201 = 603 kJ/mol
3 × O-H = 3 × 464 = 1,392 kJ/mol
The ΔHrxn for the above equation is therefore;
ΔHrxn = 1,170 + 1,004 - (603 + 1,392) = 179 kJ/mol
Answer:
The mass was there all along, it was just in the air. The weight of the oxygen from the air is not weighed in the beginning, only at the end as part of the product, making it seem like there is a total mass change.
2:7 ratio of ethane to O2 = 15:x 9solve for x)
x=52.5 mol O2