What is the limiting reactant when 45.9g of CuO are exposed to 2.57 g of H2 according to the equation
CuO(s) + H2(g)−→Cu(s) + H2O(g) ?
1.CuO
2.Unable to determine.
3.H2
4.Neither is limiting.
To determine the limiting reactant, we need to compare the amount of each reactant to the stoichiometric ratio in the balanced equation.
First, let's calculate the number of moles for each reactant:
1. CuO: molar mass of CuO = 63.55 g/mol + 16.00 g/mol = 79.55 g/mol. Therefore, moles of CuO = 45.9 g / 79.55 g/mol = 0.577 mol.
2. H2: molar mass of H2 = 2.02 g/mol. Therefore, moles of H2 = 2.57 g / 2.02 g/mol = 1.27 mol.
Next, we need to find the stoichiometric ratio for the balanced equation:
According to the equation, the ratio of CuO to H2 is 1:1. That means 1 mole of CuO reacts with 1 mole of H2.
Now, let's compare the moles of each reactant to the stoichiometric ratio:
- CuO: 0.577 mol
- H2: 1.27 mol
Since the mole ratio of CuO to H2 is 1:1, if either the moles of CuO or H2 is less than 1.27 mol, it would be the limiting reactant.
In this case, the moles of CuO (0.577 mol) is less than the moles of H2 (1.27 mol), which means CuO is the limiting reactant.
Therefore, the answer is: 1. CuO is the limiting reactant.