1.38 mol HCl and 2.9 mol NaOH react accord-
ing to the equation
HCl + NaOH −→ NaCl + H2O .
If the limiting reactant is HCl, determine
the amount of excess reactant that remains.
Answer in units of mol.
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To determine the amount of excess reactant that remains, we first need to calculate the amount of HCl and NaOH that react.
The balanced chemical equation tells us that 1 mol of HCl reacts with 1 mol of NaOH to produce 1 mol of NaCl and 1 mol of H2O.
Since the stoichiometry is 1:1, the limiting reactant is the one with fewer moles. In this case, it is HCl.
Given:
- 1.38 mol of HCl
- 2.9 mol of NaOH
To find out how many moles of NaOH react with 1.38 mol of HCl, we use the stoichiometry of the balanced equation:
1 mol HCl reacts with 1 mol NaOH.
Therefore, 1.38 mol of HCl will react with 1.38 mol of NaOH.
To determine the excess reactant, we subtract the amount of NaOH that reacted from the original amount (2.9 mol):
Excess NaOH = Original amount - Amount used
Excess NaOH = 2.9 mol - 1.38 mol
Excess NaOH = 1.52 mol
So, the amount of excess reactant (NaOH) that remains is 1.52 mol.