Question

Sodium hydroxide reacts with hydrochloric acid and water according to the equation below. Given the atomic masses, Na=23, O=16, H=1 NaOH+HCl=NaCl+H2O Which substance will be in excess if 10g of sodium hydroxide reacts with 10g of hydrochloric acid?

Answer 1

To determine which substance will be in excess, we need to calculate the number of moles of each reactant and compare their stoichiometric ratios.

First, let's calculate the number of moles for each substance:

1. Sodium hydroxide (NaOH): Moles of NaOH = mass of NaOH / molar mass of NaOH = 10 g / (23 g/mol + 16 g/mol + 1 g/mol) = 10 g / 40 g/mol = 0.25 mol

2. Hydrochloric acid (HCl): Moles of HCl = mass of HCl / molar mass of HCl = 10 g / (1 g/mol + 35.5 g/mol) = 10 g / 36.5 g/mol ≈ 0.274 mol

From the balanced equation, we can see that the stoichiometric ratio between NaOH and HCl is 1:1. This means that 1 mole of NaOH reacts with 1 mole of HCl.

Comparing the moles of NaOH and HCl, we can see that NaOH is the limiting reactant because it has fewer moles (0.25 mol) compared to HCl (approximately 0.274 mol). This indicates that HCl will be in excess.

Therefore, in the given reaction between 10 g of sodium hydroxide and 10 g of hydrochloric acid, hydrochloric acid (HCl) will be in excess.