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PLEASE URGENT ! 100 POINTS

If 16.0 g of AgNO3 react with an excess of BaCl2 according to the following equation, what mass of AgCl gets produced?

2 AgNO3+ BaCl2 → 2 AgCl +Ba(NO3)2

If only 10.0 g of AgCl were recovered, what is the percent yield of this reaction?

User BeGreen
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1 Answer

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The molar mass of AgNO3 is 169.87 g/mol, and the molar mass of AgCl is 143.32 g/mol. We can use these values to calculate the theoretical yield of AgCl:

First, we need to determine the limiting reagent in the reaction. We have 16.0 g of AgNO3, which is:

16.0 g AgNO3 x (1 mol AgNO3/169.87 g AgNO3) = 0.0942 mol AgNO3

If we assume an excess of BaCl2, we can calculate the maximum amount of AgCl that can be produced:

0.0942 mol AgNO3 x (2 mol AgCl/2 mol AgNO3) x (143.32 g AgCl/1 mol AgCl) = 26.99 g AgCl (theoretical yield)

However, only 10.0 g of AgCl were recovered. We can calculate the percent yield of the reaction as follows:

Percent yield = (actual yield/theoretical yield) x 100%

Percent yield = (10.0 g AgCl/26.99 g AgCl) x 100% = 37.05%

Therefore, the percent yield of the reaction is 37.05%.
User CMarius
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