Question

How many grams of NaCl will be produced when 2235 g of HCl are neutralized by an excess of NaOH according to the equation below?

HCl + NaOH ---> H2O + NaCl

Answer 1

Answer on Question #50322 - Chemistry - Other

Question

How many grams of NaCl will be produced when 2235 g of HCl are neutralized by an excess of NaOH according to the equation below?

$$\mathrm{HCl} + \mathrm{NaOH} \longrightarrow \mathrm{H_2O} + \mathrm{NaCl}$$

Answer:

Molar mass of NaCl equals:

$$M(\mathrm{NaCl}) = M(\mathrm{Na}) + M(\mathrm{Cl}) = 23.0 + 35.5 = 58.5 \frac{g}{\text{mole}}$$

Molar mass of HCl equals:

$$M(\mathrm{HCl}) = M(\mathrm{H}) + M(\mathrm{Cl}) = 1.0 + 35.5 = 36.5 \frac{g}{\text{mole}}$$

If 1 mole of HCl is neutralized, 1 mole of NaCl will be produced, i.e. 58.5 g of NaCl form by neutralizing 36.5 g of HCl. Then we make a proportion:

$$58.5 \text{ g of NaCl form by neutralizing } 36.5 \text{ g of HCl}$$ $$x \text{ g of NaCl} - 2235 \text{ g of HCl}$$ $$x = \frac{58.5 \cdot 2235}{36.5} = 3582.1 \text{ g}$$

Answer: $m(\mathrm{NaCl}) = 3582.1 \text{ g}$

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