1. Ten moles of N
2O
4(l) are added to an unspecified amount N
23(CH
3)(l) according to the equation shown below:
5 N
2O
4(l) + 4 N
2H
3(CH
3(l)
12 H
2O(g) + N
2 + CO
2(g)
If 23 moles of water are produced and the reaction runs to completion, what is the limiting reagent?
A) N
2S
4(l)
B) N
2H
3(CH
3)(l)
C) H
2O(g)
D) There is no limiting reagent.
I ended up getting an answer of C for this one, thinking that we could use the number of moles of H
2O and moles of N
2O
4 from the reaction to determine the amount of total N
2H
3 produced. However, in reviewing the answering for it, the rationale was that the correct answer was B since we weren't provided with a specified amount and that the reaction would have produced a total of 24 mol of H
20 with N
2O but how can we assume this to be the case if we're not given a starting amount of N
2H
3?
2. The following reaction is run to completion:
As
4O
6(s) + 6C(s)
As
4(g) + 6CO(g)
How much carbon is required to produce 75 g of arsenic acid?
A) 12 g
B) 18 g
C) 48 g
D) 72 g
This is how I went about trying to solve it:
75 g As x (1 mol As/300 g As) x (6 mol C/1 mol As) x (12 g C/1 mol C).
I'm not sure if this is the correct approach or not?
Any help is appreciated!