In the following equations, determined which reactant is the limiting reactant and which reactant is in excess. The amount used are given below. KOH+HNO3->KNO3+H2O 15.4kg KOH and 12.0g HNO3 react.
To determine the limiting reactant, we must compare the number of moles of each reactant.
First, we need to convert the given mass of each reactant to moles. The molar mass of KOH is 56.11 g/mol, and the molar mass of HNO3 is 63.01 g/mol.
Moles of KOH = mass of KOH / molar mass of KOH
Moles of KOH = 15,400 g / 56.11 g/mol
Moles of KOH = 274.77 mol
Moles of HNO3 = mass of HNO3 / molar mass of HNO3
Moles of HNO3 = 12.0 g / 63.01 g/mol
Moles of HNO3 = 0.19 mol
Now, we can compare the moles of each reactant:
Moles of KOH = 274.77
Moles of HNO3 = 0.19
We see that the moles of HNO3 are significantly lower than the moles of KOH. Therefore, HNO3 is the limiting reactant.
To determine the excess reactant, we subtract the moles of the limiting reactant from the moles of the other reactant:
Excess moles of KOH = Moles of KOH - Moles of HNO3
Excess moles of KOH = 274.77 - 0.19
Excess moles of KOH = 274.58
Since the result is positive, it means that there is excess KOH.