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For the reaction shown, calculate the number of grams of product formed from 56.7 grams of iron and sufficient chlorine. Enter your answer with the correct significant figures. Do not include units with your answer.

[tex]\[ 2 \text{Fe} (s) + 3 \text{Cl}_2(g) \rightarrow 2 \text{FeCl}_3(s) \][/tex]

[tex]\[\boxed{\phantom{answer}}\][/tex]


Sagot :

To determine the number of grams of iron(III) chloride ([tex]\(\text{FeCl}_3\)[/tex]) produced from 56.7 grams of iron (Fe) with sufficient chlorine (Cl[tex]\(_2\)[/tex]), follow these steps:

1. Identify the Molar Masses:
- Molar mass of iron ([tex]\(\text{Fe}\)[/tex]): 55.845 g/mol.
- Molar mass of iron(III) chloride ([tex]\(\text{FeCl}_3\)[/tex]): 162.204 g/mol.

2. Calculate the Number of Moles of Iron:
[tex]\[ \text{Moles of Fe} = \frac{\text{Mass of Fe}}{\text{Molar Mass of Fe}} = \frac{56.7 \, \text{g}}{55.845 \, \text{g/mol}} \approx 1.0153 \, \text{moles} \][/tex]

3. Determine the Mole Ratio:
From the balanced chemical equation:
[tex]\[ 2 \, \text{Fe} + 3 \, \text{Cl}_2 \rightarrow 2 \, \text{FeCl}_3 \][/tex]
The mole ratio of [tex]\(\text{Fe}\)[/tex] to [tex]\(\text{FeCl}_3\)[/tex] is 1:1. Therefore, the moles of [tex]\(\text{FeCl}_3\)[/tex] produced will be the same as the moles of [tex]\(\text{Fe}\)[/tex] used:
[tex]\[ \text{Moles of FeCl}_3 = 1.0153 \, \text{moles} \][/tex]

4. Calculate the Mass of [tex]\(\text{FeCl}_3\)[/tex]:
[tex]\[ \text{Mass of FeCl}_3 = \text{Moles of FeCl}_3 \times \text{Molar Mass of FeCl}_3 = 1.0153 \, \text{moles} \times 162.204 \, \text{g/mol} \approx 164.687 \, \text{grams} \][/tex]

5. Adjust for Significant Figures:
The given mass of iron (56.7 g) has three significant figures. Therefore, the final mass of [tex]\(\text{FeCl}_3\)[/tex] should also be reported with three significant figures:
[tex]\[ \text{Mass of FeCl}_3 = 164.687 \, \text{grams} \][/tex]

To conclude, the mass of [tex]\(\text{FeCl}_3\)[/tex] produced from 56.7 grams of iron is approximately [tex]\(164.687\)[/tex] grams.