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Sagot :
Sure, let's solve this step-by-step.
1. Determine the balanced chemical equation:
The balanced equation is:
[tex]\[ \mathrm{Fe_2O_3 + 3H_2 \rightarrow 2Fe + 3H_2O} \][/tex]
2. Identify given information:
- Mass of [tex]\( \mathrm{H_2} \)[/tex]: [tex]\( 5.26 \)[/tex] grams
- Molar mass of [tex]\( \mathrm{H_2} \)[/tex]: [tex]\( 2.016 \, \text{g/mol} \)[/tex]
- Molar mass of [tex]\( \mathrm{Fe} \)[/tex]: [tex]\( 55.845 \, \text{g/mol} \)[/tex]
3. Calculate the number of moles of [tex]\( \mathrm{H_2} \)[/tex]:
To find the number of moles of [tex]\( \mathrm{H_2} \)[/tex], use the formula:
[tex]\[ \text{Moles of } \mathrm{H_2} = \frac{\text{Mass of } \mathrm{H_2}}{\text{Molar mass of } \mathrm{H_2}} \][/tex]
[tex]\[ \text{Moles of } \mathrm{H_2} = \frac{5.26 \, \text{grams}}{2.016 \, \text{g/mol}} \][/tex]
[tex]\[ \text{Moles of } \mathrm{H_2} = 2.609126984126984 \, \text{mol} \][/tex]
4. Using stoichiometry to find moles of [tex]\( \mathrm{Fe} \)[/tex]:
From the balanced equation, 3 moles of [tex]\( \mathrm{H_2} \)[/tex] produce 2 moles of [tex]\( \mathrm{Fe} \)[/tex]. Thus, we need to find out how many moles of [tex]\( \mathrm{Fe} \)[/tex] can be produced by 2.609126984126984 moles of [tex]\( \mathrm{H_2} \)[/tex]:
[tex]\[ \text{Moles of } \mathrm{Fe} = \left( \frac{2 \, \text{moles of } \mathrm{Fe}}{3 \, \text{moles of } \mathrm{H_2}} \right) \times \text{Moles of } \mathrm{H_2} \][/tex]
[tex]\[ \text{Moles of } \mathrm{Fe} = \left( \frac{2}{3} \right) \times 2.609126984126984 \][/tex]
[tex]\[ \text{Moles of } \mathrm{Fe} = 1.7394179894179895 \, \text{mol} \][/tex]
5. Calculate the mass of [tex]\( \mathrm{Fe} \)[/tex] produced:
To find the mass of [tex]\( \mathrm{Fe} \)[/tex] produced, use the formula:
[tex]\[ \text{Mass of } \mathrm{Fe} = \text{Moles of } \mathrm{Fe} \times \text{Molar mass of } \mathrm{Fe} \][/tex]
[tex]\[ \text{Mass of } \mathrm{Fe} = 1.7394179894179895 \times 55.845 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of } \mathrm{Fe} = 97.13779761904762 \, \text{grams} \][/tex]
Therefore, from 5.26 grams of [tex]\( \mathrm{H_2} \)[/tex], we can produce approximately 97.14 grams of iron ([tex]\( \mathrm{Fe} \)[/tex]).
1. Determine the balanced chemical equation:
The balanced equation is:
[tex]\[ \mathrm{Fe_2O_3 + 3H_2 \rightarrow 2Fe + 3H_2O} \][/tex]
2. Identify given information:
- Mass of [tex]\( \mathrm{H_2} \)[/tex]: [tex]\( 5.26 \)[/tex] grams
- Molar mass of [tex]\( \mathrm{H_2} \)[/tex]: [tex]\( 2.016 \, \text{g/mol} \)[/tex]
- Molar mass of [tex]\( \mathrm{Fe} \)[/tex]: [tex]\( 55.845 \, \text{g/mol} \)[/tex]
3. Calculate the number of moles of [tex]\( \mathrm{H_2} \)[/tex]:
To find the number of moles of [tex]\( \mathrm{H_2} \)[/tex], use the formula:
[tex]\[ \text{Moles of } \mathrm{H_2} = \frac{\text{Mass of } \mathrm{H_2}}{\text{Molar mass of } \mathrm{H_2}} \][/tex]
[tex]\[ \text{Moles of } \mathrm{H_2} = \frac{5.26 \, \text{grams}}{2.016 \, \text{g/mol}} \][/tex]
[tex]\[ \text{Moles of } \mathrm{H_2} = 2.609126984126984 \, \text{mol} \][/tex]
4. Using stoichiometry to find moles of [tex]\( \mathrm{Fe} \)[/tex]:
From the balanced equation, 3 moles of [tex]\( \mathrm{H_2} \)[/tex] produce 2 moles of [tex]\( \mathrm{Fe} \)[/tex]. Thus, we need to find out how many moles of [tex]\( \mathrm{Fe} \)[/tex] can be produced by 2.609126984126984 moles of [tex]\( \mathrm{H_2} \)[/tex]:
[tex]\[ \text{Moles of } \mathrm{Fe} = \left( \frac{2 \, \text{moles of } \mathrm{Fe}}{3 \, \text{moles of } \mathrm{H_2}} \right) \times \text{Moles of } \mathrm{H_2} \][/tex]
[tex]\[ \text{Moles of } \mathrm{Fe} = \left( \frac{2}{3} \right) \times 2.609126984126984 \][/tex]
[tex]\[ \text{Moles of } \mathrm{Fe} = 1.7394179894179895 \, \text{mol} \][/tex]
5. Calculate the mass of [tex]\( \mathrm{Fe} \)[/tex] produced:
To find the mass of [tex]\( \mathrm{Fe} \)[/tex] produced, use the formula:
[tex]\[ \text{Mass of } \mathrm{Fe} = \text{Moles of } \mathrm{Fe} \times \text{Molar mass of } \mathrm{Fe} \][/tex]
[tex]\[ \text{Mass of } \mathrm{Fe} = 1.7394179894179895 \times 55.845 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of } \mathrm{Fe} = 97.13779761904762 \, \text{grams} \][/tex]
Therefore, from 5.26 grams of [tex]\( \mathrm{H_2} \)[/tex], we can produce approximately 97.14 grams of iron ([tex]\( \mathrm{Fe} \)[/tex]).
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