To solve this problem we need to calculate the molar mas of Cu and CuO:
[tex]\begin{gathered} Mm_{Cu}=63.546\text{ g/mol} \\ Mm_{CuO}=79.545\text{ g/mol} \end{gathered}[/tex]because of the chemical equation we know that 2 mols of Cu produce 2 mols of CuO. Now we have to convert those mols into mass, to do that we multiply the number of mols by the molar mass already calculated:
[tex]m_{CuO}=n_{CuO}\cdot Mm_{CuO}=2mol\cdot79.545g/mol=159.09g[/tex]this means that for 2 mols of Cu that are consumed 159.09 g of CuO are produced. Then we just need to use this information to calculate the right proportion for 2.87mols of Cu. we can do that using a conversion factor:
[tex]2.87\text{ mols of Cu}\cdot\frac{159.09\text{ g of CuO}}{2mols\text{ of Cu}}=228.294g\text{ of CuO}[/tex]228.294 g of CuO are produced
