mols Cu = grams/atomic mass = 8.75/63.5 = ?
mols NO produced = mols Cu x (2 mols NO/3 mols Cu) = ?
Then you know 1 mol NO occupies a volume of 22.4 L @ stp
What volume of dry NO(g) at STP could be produced by reacting 8.74 g of Cu with an excess of HNO3?
3Cu + 8 HNO3→ 3 Cu(NO3)2 + 2 NO(g) + 4 H2O(l)
2 answers
Thank you, this really helped
1 answer