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For the following reaction, 4.57 g of silver nitrate are mixed with excess copper. The reaction yields 2.29 gram of copper(II) nitrate What is the percent yield for this reaction?Formula: % yield = (Actual yield/theoretical yield) x 100 2 AgNO3(aq) + Cu(s) à Cu(NO3)2 (aq) + 2 Ag(s)

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sebassandin

Answer:

[tex]Y=90.6\%[/tex]

Explanation:

Hello,

In this case, for the given chemical reaction:

[tex]2AgNO_3(aq) + Cu(s) \rightarrow Cu(NO_3)_2 (aq) + 2 Ag(s)[/tex]

If 4.57 grams of silver nitrate are used in copper excess, the theoretical yield of copper (II) nitrate turns out:

[tex]m_{Cu(NO_3)_2}^{theoretical}=4.57gAgNO_3*\frac{1molAgNO_3}{170gAgNO_3}*\frac{1molCu(NO_3)_2}{2molAgNO_3}*\frac{188gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}^{theoretical}=2.53gCu(NO_3)_2[/tex]

In such a way, the percent yield results:

[tex]Y=\frac{2.29g}{2.53g}*100\%=90.6\%[/tex]

Best regards.

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