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Math  /  Algebra

Question4HCl( g)+O2( g)→2Cl2( g)+2H2O( g)4 \mathrm{HCl}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g})
Calculate the number of grams of Cl2\mathrm{Cl}_{2} formed when 0.225 mol HCl reacts with an excess of O2\mathrm{O}_{2}. mass: □\square g\mathrm{g}

Studdy Solution

STEP 1

1. The chemical reaction is given by the equation: 4HCl( g)+O2( g)→2Cl2( g)+2H2O( g)4 \mathrm{HCl}(\mathrm{~g}) + \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Cl}_{2}(\mathrm{~g}) + 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}).
2. We have 0.225 moles of HCl\mathrm{HCl}.
3. O2\mathrm{O}_{2} is in excess, so it will not limit the reaction.
4. We need to calculate the mass of Cl2\mathrm{Cl}_{2} produced.

STEP 2

1. Determine the mole ratio from the balanced equation.
2. Calculate the moles of Cl2\mathrm{Cl}_{2} produced.
3. Convert moles of Cl2\mathrm{Cl}_{2} to grams.

STEP 3

Determine the mole ratio from the balanced equation.
From the balanced equation, 4 moles of HCl\mathrm{HCl} produce 2 moles of Cl2\mathrm{Cl}_{2}.

STEP 4

Calculate the moles of Cl2\mathrm{Cl}_{2} produced.
Using the mole ratio, calculate the moles of Cl2\mathrm{Cl}_{2} produced from 0.225 moles of HCl\mathrm{HCl}:
Moles of Cl2=0.225 mol HCl×2 mol Cl24 mol HCl\text{Moles of } \mathrm{Cl}_{2} = 0.225 \, \text{mol HCl} \times \frac{2 \, \text{mol Cl}_{2}}{4 \, \text{mol HCl}}
Moles of Cl2=0.1125 mol Cl2\text{Moles of } \mathrm{Cl}_{2} = 0.1125 \, \text{mol Cl}_{2}

STEP 5

Convert moles of Cl2\mathrm{Cl}_{2} to grams.
The molar mass of Cl2\mathrm{Cl}_{2} is approximately 70.90 g/mol70.90 \, \text{g/mol}.
Calculate the mass:
Mass of Cl2=0.1125 mol Cl2×70.90 g/mol\text{Mass of } \mathrm{Cl}_{2} = 0.1125 \, \text{mol Cl}_{2} \times 70.90 \, \text{g/mol}
Mass of Cl2=7.978125 g\text{Mass of } \mathrm{Cl}_{2} = 7.978125 \, \text{g}
Rounding to three significant figures, the mass of Cl2\mathrm{Cl}_{2} is:
7.98 g\boxed{7.98 \, \text{g}}

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