3.00 g of CUO and 1.00 g of ale are combined in-law to react give, which one is limiting ?
To determine which reactant is limiting, we must first convert the mass of each reactant to moles.
1. Calculate the moles of CuO:
Molar mass of CuO (Cu = 63.55 g/mol, O = 16 g/mol)
Molar mass of CuO = 63.55 g/mol + 16 g/mol = 79.55 g/mol
Moles of CuO = 3.00 g / 79.55 g/mol ≈ 0.038 moles
2. Calculate the moles of Al:
Molar mass of Al = 26.98 g/mol
Moles of Al = 1.00 g / 26.98 g/mol ≈ 0.037 moles
Next, we need to determine the stoichiometry of the reaction between CuO and Al. The balanced chemical equation for the reaction is:
3CuO + 2Al → 3Cu + Al2O3
From the balanced equation, we can see that 3 moles of CuO react with 2 moles of Al.
Now, we need to determine how many moles of Al would be required to completely react with the moles of CuO present:
0.038 moles of CuO / 3 × (2 moles of Al / 3 moles of CuO) = 0.025 moles of Al
Since we only have 0.037 moles of Al available, Al is the limiting reactant in this reaction.