yes and yes.
8HNO3 + 3Cu ==>
mols HNO3 initially = M x L = ?
mols Cu = grams/atomic mass = ?
Now calculate how much HNO3 is need to dissolve that many mols Cu.
mols HNO3 remaining unreacted = initial mols HNO3 = mols needed to dissolve the Du = ?
A student performing an experiment mistakenly used 8.0 ml of 16 M HNO3 to dissolve 0.16g of solid copper instead of the 4.0 ml as suggested. how many moles of HNO3 remains after the reaction with copper is complete. 8 moles of HNO3 reacts with 3 moles of Cu
im not sure how i would start this equation. would i start from the moles of HNO3 calcualted from the 8ml of 16 M HNO3 or with the .16 g of Cu
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