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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)

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change 120 ml to l
.120 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

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