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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%

Respuesta :

sebassandin

Answer:

D. 91.0%

Explanation:

Hello,

In this case, for the given chemical reaction:

[tex]Na_2S+2AgNO_3 \rightarrow Ag_2S + 2NaNO_3[/tex]

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:

[tex]m_{Ag_2S}^{theoretical}=4.27gAgNO_3*\frac{1molAgNO_3}{169.87gAgNO_3} *\frac{1molAg_2S}{2molAgNO_3}*\frac{247.8gAg_2S}{1molAg_2S}\\\\m_{Ag_2S}^{theoretical}=3.11gAg_2S[/tex]

Next, we compute the percent yield as:

[tex]Y=\frac{m_{Ag_2S}^{actual}}{m_{Ag_2S}^{theoretical}}*100\% =\frac{2.86g}{3.11g} *100\%\\\\Y=91.0%[/tex]

Hence, answer is D. 91.0%.

Best regards.

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