Question

0,20 dm3 of 0,20 moldm3 hydrochloric acid were reacted with excess calcium.what volume of hydrogen was produced

Answer 1

To determine the volume of hydrogen gas produced when hydrochloric acid (HCl) reacts with excess calcium (Ca), we can start by writing the balanced chemical equation for the reaction:

\[
\text{Ca (s) + 2 HCl (aq) } \rightarrow \text{ CaCl}_2 \text{ (aq) + H}_2 \text{ (g)}
\]

From this equation, we see that 1 mole of calcium reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas.

### Step 1: Calculate moles of HCl

We have a \(\text{0.20 dm}^3\) (or 200 mL) solution of \(\text{0.20 mol/dm}^3\) HCl.

\[
\text{Moles of HCl} = \text{Volume (dm}^3\text{) } \times \text{ Concentration (mol/dm}^3\text{)}
\]
\[
\text{Moles of HCl} = 0.20 \, \text{dm}^3 \times 0.20 \, \text{mol/dm}^3 = 0.04 \, \text{moles}
\]

### Step 2: Moles of Hydrogen Produced

From the reaction stoichiometry, 2 moles of HCl produce 1 mole of H2. Therefore, the moles of hydrogen produced from 0.04 moles of HCl can be calculated as follows:

\[
\text{Moles of } H_2 = \frac{0.04 \, \text{moles HCl}}{2} = 0.02 \, \text{moles of } H_2
\]

### Step 3: Volume of Hydrogen Gas produced

Using the ideal gas law, at standard temperature and pressure (STP, which is 0°C and 1 atm), 1 mole of gas occupies 22.4 dm³.

\[
\text{Volume of } H_2 = \text{Moles of } H_2 \times \text{Volume per mole at STP}
\]
\[
\text{Volume of } H_2 = 0.02 \, \text{moles} \times 22.4 \, \text{dm}^3/\text{mol} = 0.448 \, \text{dm}^3
\]

### Conclusion

The volume of hydrogen gas produced is **0.448 dm³** (or 448 mL).