posted by helen on .
How many g NO gas can be produced from 120 mL of 2.80 M HNO3 and excess Cu?
3Cu(s) + 8HNO3(aq) ---> 3Cu (NO3)2(aq) + 4H2O(l) + 2NO(g)
change 120 ml to l
then divide M by L to get moles
2.80 M/.120 L
then take this and multiply it by the atomic weight or mass of NO