if 41.6 g of N(2)O(4) reacts with 20.8 g of N(2)H(4) the products will be nitrogen and water. What mass of water will be produced?
First you need to balance the chemical reaction. The key is to noticing that there is twice as much hydrogen as oxygen in water.
N2O4 + 2(N2H4) -> 3(N2) + 4(H2O)
From here it becomes a limiting reagent problem.
Add up the mass of both reactants.
Find the amounts of each reagent by dividing the given amount with its molecular mess
Review the original reaction to see that N2O2 is 1:1 and would provide enough oxygen to produce 0.465 moles of water. N2H4 however is 2:1 and therefore needs to adjusted 0.649/2=0.324 moles of water.
meaning that N2H4 is the limiting reagent.
We take the moles of water produced and multiply it by the molecular weight of water.
Molecular weight water:
0.324*18.015=5.837g of water