In the following equations, determined which reactant is the limiting reactant and which reactant is in excess. The amount used are given below.
KO+HNO3-> KNO3+H2O 15.4 KOH and HNO3 react.
To determine the limiting reactant, we need to compare the moles of each reactant used to the stoichiometric ratio given by the balanced equation.
The balanced equation is:
2 KOH + HNO3 -> KNO3 + H2O
From the equation, we can see that 2 moles of KOH react with 1 mole of HNO3 to produce 1 mole of KNO3 and 1 mole of H2O.
First, we need to convert the given amounts in grams to moles.
The molar mass of KOH is 56.11 g/mol (from periodic table)
Moles of KOH = 15.4 g / 56.11 g/mol = 0.275 moles
The molar mass of HNO3 is 63.01 g/mol (from periodic table)
Moles of HNO3 = 15.4 g / 63.01 g/mol = 0.244 moles
Now, we can compare the moles of each reactant to the stoichiometric ratio.
The stoichiometric ratio tells us that 2 moles of KOH react with 1 mole of HNO3.
The ratio of moles of KOH to HNO3 is:
0.275 moles KOH / 0.244 moles HNO3 ≈ 1.13
Since the ratio is greater than 1, it means that there is more KOH than HNO3. Therefore, KOH is in excess and HNO3 is the limiting reactant.