Question

For the following reaction, if 17.8 g of Na2S, is reacted with excess AgNO3 in the laboratory, and 31 g of NaNO3 is produced, what is the percentage yield of NaNO3? Round your answer to the nearest whole percent.Na2S + 2 AgNO3 → Ag2S + 2 NaNO3Select one:a.26 %b.92 %c.75 %d.80 %

Answer 1

In this question, we have the following reaction occurring:

Na2S + 2 AgNO3 --> Ag2S + 2 NaNO3

We have:

17.8 grams of Na2S

31 grams of NaNO3 actually produced

Now we have to find the theoretical yield of NaNO3 that would be produced from 17.8 grams of Na2S

Using the molar mass of Na2S, 78.04g/mol, we can find the number of moles:

78.04 grams = 1 mol

17.8 grams = x moles

78.04x = 17.8

x = 17.8/78.04

x = 0.23 moles of Na2S in 17.8 grams

The molar ratio between Na2S and NaNO3 is 1:2, therefore we will produce two times more moles of NaNO3 than Na2S

0.23 * 2 = 0.46 moles of NaNO3 will be produced

Using the molar mass of NaNO3, 85g/mol, we can find the mass:

85 grams = 1 mol

x grams = 0.46 moles

x = 0.46 * 85

x = 39.1 grams

Now to find the percent yield, we will use the following formula:

%yield = actual yield/theoretical yield

%yield = 31/39.1

%yield = 0.8 or 80%

The percent yield will be 80% letter D