According to the reaction NaOH + HCl �¨ NaCl + H2O, which is the limiting reactant when 37.0 g of HCl are mixed with 37.0 g NaOH?

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To determine the limiting reactant in a chemical reaction, you need to compare the number of moles of each reactant and their stoichiometric ratios with their balanced equation.

First, let's determine the number of moles of HCl and NaOH using their respective molar masses.

The molar mass of HCl (hydrochloric acid) is:
H = 1.01 g/mol
Cl = 35.45 g/mol
So, the molar mass of HCl is 1.01 + 35.45 = 36.46 g/mol.

To find the number of moles of HCl, divide the mass (37.0 g) by the molar mass:
Number of moles of HCl = 37.0 g / 36.46 g/mol ≈ 1.014 mol

The molar mass of NaOH (sodium hydroxide) is:
Na = 22.99 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
So, the molar mass of NaOH is 22.99 + 16.00 + 1.01 = 40.00 g/mol.

To find the number of moles of NaOH, divide the mass (37.0 g) by the molar mass:
Number of moles of NaOH = 37.0 g / 40.00 g/mol ≈ 0.925 mol

Now, let's compare the mole ratios of HCl and NaOH in the balanced equation:
1 mol of HCl reacts with 1 mol of NaOH.

Based on the balanced equation and the given moles of reactants:
HCl:NaOH = 1.014 mol:0.925 mol

Since the stoichiometric ratio of HCl:NaOH is 1:1, we can see that HCl is in excess compared to NaOH. This means NaOH is the limiting reactant because it will be completely consumed in the reaction while some HCl will be left unreacted.

Therefore, in the given reaction, NaOH is the limiting reactant when 37.0 g of HCl is mixed with 37.0 g NaOH.

Convert HCl to mols.

Convert NaOH to mols.
Convert mols HCl to mols NaCl.
Convert mols NaOH to mols NaCl.
The smaller value is the actual mols NaCl formed and the reagent providing that value is the limiting reagent.