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4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment. At the end of the experiment he obtained 0.038 g of a black product. What was his percent yield

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sebassandin

Answer:

[tex]Y=48.6\%[/tex]

Explanation:

Hello,

In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:

[tex]2Cu+O_2\rightarrow 2CuO[/tex]

In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:

[tex]m_{CuO}=0.032gCu*\frac{1molCu}{63.546gCu} *\frac{2molCuO}{2molCu}*\frac{79.546gCuO}{1molCuO} =0.078gCuO[/tex]

Therefore, the percent yield is:

[tex]Y=\frac{0.038g}{0.078g}*100\%\\ \\Y=48.6\%[/tex]

Best regards.

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