If 25.0 gN, and 18.75 g H, are used for this reaction, which reactant will be the limiting reactant?
0) How much of the excess resctant will be left over?
To determine which reactant is the limiting reactant, we need to calculate the moles of each reactant and then use the stoichiometry of the balanced chemical equation to see which reactant will run out first.
1. Calculate the moles of N:
molar mass of N = 14.01 g/mol
moles of N = 25.0 g / 14.01 g/mol = 1.782 mol
2. Calculate the moles of H:
molar mass of H = 1.01 g/mol
moles of H = 18.75 g / 1.01 g/mol = 18.564 mol
3. Now we need to determine which reactant will be the limiting reactant by comparing the moles of each reactant to the stoichiometry of the balanced chemical equation.
The balanced chemical equation for the reaction isn't provided, so we have to assume it's a simple combination of N and H in a 1:1 ratio. If that's the case, N will be the limiting reactant because it has fewer moles (1.782 mol) than H (18.564 mol).
4. To find out how much excess H will be left over, we can use the stoichiometry of the balanced chemical equation. In this case, since N is the limiting reactant, all of the 1.782 mol of N will react with 1.782 mol of H. That means there will be an excess of (18.564 - 1.782) mol = 16.782 mol of H left over.
0) The excess reactant in this case is hydrogen, and there will be 16.782 mol of hydrogen left over after the reaction is complete.