Question

Consider the reaction MnO2 + 4HCl = MnCl2 + Cl2 + 2 H2O. If 0.86 moles of MnO2 and 48.2 g of HCl react, how many grams of Cl2 will be produced?

a) 35.5 g
b) 71.0 g
c) 106.5 g
d) 53.25 g

Answer 1

The grams of Cl2 produced in the reaction will be approximately 71.0 g (Option b).

Step-by-step explanation:

In order to determine the grams of Cl2 produced, we first need to identify the limiting reactant. Convert 0.86 moles of MnO2 to moles of Cl2 using the stoichiometric coefficients from the balanced equation. Then, convert grams of HCl to moles of Cl2. The smaller value obtained is the limiting reactant. Use this information to find the moles of Cl2 produced and finally convert it to grams.

Firstly, convert 0.86 moles of MnO2 to moles of Cl2 using the balanced equation:

`\[ MnO2 + 4HCl \rightarrow MnCl2 + Cl2 + 2H2O \]`

From the equation, 1 mole of MnO2 produces 1 mole of Cl2. Therefore, 0.86 moles of MnO2 produce 0.86 moles of Cl2.

Next, convert 48.2 g of HCl to moles of Cl2 using the molar mass of HCl:

`\[ \text{Molar mass of HCl} = 1.01 \, \text{g/mol} + 35.45 \, \text{g/mol} = 36.46 \, \text{g/mol} \]`

`\[ \text{Moles of HCl} = \frac{48.2 \, \text{g}}{36.46 \, \text{g/mol}} \approx 1.32 \, \text{moles of HCl} \]`

Now, compare the moles of Cl2 obtained from MnO2 (0.86 moles) and HCl (1.32 moles). The limiting reactant is MnO2. Therefore, 0.86 moles of Cl2 will be produced.

Finally, convert moles of Cl2 to grams using the molar mass of Cl2:

`\[ \text{Molar mass of Cl2} = 2 * 35.45 \, \text{g/mol} = 70.90 \, \text{g/mol} \]`

`\[ \text{Grams of Cl2 produced} = 0.86 \, \text{moles} * 70.90 \, \text{g/mol} \approx 71.0 \, \text{g} \]`

Therefore, the correct answer is approximately 71.0 g (Option b).