Question

What is the maximum number of grams of copper that could be produced by the reaction of 30.0 of copper oxide with excess methane? CuO(s)+CH4(I)-H2O(I)+Cu(s)+CO2(g)

Answer 1

Answer: 24.13 g Cu

Explanation:

Given for this question:

M of CuO = 30 g

m of CuO = 79.5 g/mol

Number of moles of CuO = (given mass ÷ molar mass) = (30 ÷ 79.5) mol

= 0.38 mol

The max number of CuO (s) that can be produced by the reaction of excess methane can be solved with this reaction:

CuO(s) + CH4(l) ------> H2O(l) + Cu(s) + CO2(g)

The balanced equation can be obtained by placing coefficients as needed and making sure the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side

4CuO(s) + CH4(l) ----> 2H2O(l) + 4Cu(s) + CO2(g)          

From the stoichiometry of the balanced equation:

4 moles of CuO gives 4 moles of Cu

1 mole of CuO gives 1 mol of Cu

0.38 mol of CuO gives 0.38 mol of Cu

Therefore, the grams of Cu that can be produced = 0.38 × molar mass of Cu

= 0.38 × 63.5 g

= 24.13 grams        

Therefore, 24.13 grams of copper could be produced by the reaction of 30.0 of copper oxide with excess methane