Question

Na2S+2AgNO3 = Ag2S + NaNO3 If 2.86g of Ag2S are actually produced by a reaction between an excess of Na2S and 4.27g of AgNO3 then what is the percent yield of Ag2S

A. 3.15%
B. 45.9%
C. 61.2%
D. 91.0%

Answer 1

D. 91.0%

Step-by-step explanation:

Hello,

In this case, for the given chemical reaction:

`Na_2S+2AgNO_3 \rightarrow Ag_2S + 2NaNO_3`

Next, since silver nitrate (molar mass 169.87 g/mol) is in a 2:1 molar ratio with silver sulfide (molar mass 247.8 g/mol), we compute its theoretical yield as shown below:

`m_(Ag_2S)^(theoretical)=4.27gAgNO_3*(1molAgNO_3)/(169.87gAgNO_3) *(1molAg_2S)/(2molAgNO_3)*(247.8gAg_2S)/(1molAg_2S)\\\\m_(Ag_2S)^(theoretical)=3.11gAg_2S`

Next, we compute the percent yield as:

`Y=(m_(Ag_2S)^(actual))/(m_(Ag_2S)^(theoretical))*100\% =(2.86g)/(3.11g) *100\%\\\\Y=91.0%`

Hence, answer is D. 91.0%.

Best regards.