Question

3Cuo+2NH3= 3Cu+3H2O+N2 with 23.6g of cuo and 6.4g of 2NH3 which reactant is limiting reactant and how many grams of cu will be produced ||| O​

Answer 1

CuO is the limiting reactant in the given reaction. Approximately 29.9 grams of Cu will be produced.

Step-by-step explanation:

In the given reaction, 3CuO + 2NH3 = 3Cu + 3H2O + N2, the first step is to determine the limiting reactant. To do this, we need to calculate the number of moles of each reactant.

For CuO: 23.6 g * (1 mol/79.55 g) = 0.297 mol

For NH3: 6.4 g * (1 mol/17.03 g) = 0.376 mol

Since the stoichiometry ratio between CuO and NH3 is 3:2, CuO is the limiting reactant because it produces less moles of Cu compared to NH3.

To find the grams of Cu produced, we can use the stoichiometry of the reaction. Since 3 mol of CuO produces 3 mol of Cu, we can set up a ratio:

0.297 mol CuO * (3 mol Cu/3 mol CuO) * (63.55 g Cu/1 mol Cu) = 29.9 g Cu