Answered

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You are performing a chemical reaction using the following equation:
2 AgI + Na2S → Ag2S + 2 NaI
You realize that you're running low on both of your reagents. You have 200.0 g each of AgI and of Na2S. What is the maximum amount of Ag2S that can be produced?

Sagot :

ihlester

Answer: max Ag2S 159 g

Explanation:

at wts Ag 107.87, I 126.9, Na 22.99, S 32.065

AgI 234.76, Na2S 87.12, Ag2S 172.0, NaI 149.6

2(mol)AgI + 1(mol)Na2S —> 1(mol)Ag2S + 2(mol)NaI

200g AgI = 200/107.87 mol = 1.85 mol

200g Na2S = 200/87.12 mol = 2.3 mol, so AgI is limiting.

2 mol AgI —> 1 mol Ag2S, so

1.85 mol AgI —> 1.85/2 mol Ag2S

1 mol Ag2S = 172 g

max Ag2S = 172*1.85/2 g = 159g

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