Question

Copper is commonly mined as an ore with a variable percent composition of copper (II) sulfide. This ore is also sometimes referred to as covellite. If a sample of the ore containing 12.5% covellite is exposed to oxygen under high temperatures, copper (II) oxide and sulfur dioxide will form. What is the yield of copper (II) oxide if 1.00 kg of the original ore is processed and the expected yield is 90%

Answer 1

Answer:

$m_{CuO}=93.6gCuO$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$CuS+\frac{3}{2} O_2\rightarrow CuO+SO_2$

Thus, given the 1.00-kg of 12.5% ore, we can compute the theoretical yield of copper (II) oxide via stoichiometry:

$m_{CuO}^{theoretical}=1.00kgCuS*\frac{1000gCuS}{1kgCuS} *\frac{12.5gCuS}{100gCuS} *\frac{1molCuS}{95.6gCuS} *\frac{1molCuO}{1molCuS} *\frac{79.5gCuO}{1molCuO} \\\\m_{CuO}^{theoretical}=103.95gCuO$

Whereas the third factor accounts for the percent purity of the covellite. Then, given the percent yield, we can compute the actual yield by:

$m_{CuO}=103.95gCuO*0.9\\\\m_{CuO}=93.6gCuO$

Regards.