Q.51. The mass of iron converted into FeO by the action of 18 g of steam is : (Given : Molar mass of

Q: What is the correct answer for this exam Chemistry Q51?

The correct answer is D. The balanced chemical equation for the reaction is: $3\mathrm{Fe(s)} + 4\mathrm{H_2O(g)} \rightarrow \mathrm{Fe_3O_4(s)} + 4\mathrm{H_2(g)}$. Molar mass of H$_2$O = $2 \times 1 + 16 = 18$ g/mol. Molar mass of Fe = 56 g/mol. Given 18 g of steam (H$_2$O). Number of moles of H$_2$O = $\frac{18 \text{ g}}{18 \text{ g/mol}} = 1$ mol. From the stoichiometry, 4 moles of H$_2$O react with 3 moles of Fe. S

Q: Q.51. The mass of iron converted into FeO by the action of 18 g of steam is : (Given : Molar mass of H, O and Fe are 1, 16 and 56 g mol respectively) Assume iron is present in excess :

The correct answer is D. The balanced chemical equation for the reaction is: $3\mathrm{Fe(s)} + 4\mathrm{H_2O(g)} \rightarrow \mathrm{Fe_3O_4(s)} + 4\mathrm{H_2(g)}$. Molar mass of H$_2$O = $2 \times 1 + 16 = 18$ g/mol. Molar mass of Fe = 56 g/mol. Given 18 g of steam (H$_2$O). Number of moles of H$_2$O = $\frac{18 \text{ g}}{18 \text{ g/mol}} = 1$ mol. From the stoichiometry, 4 moles of H$_2$O react with 3 moles of Fe. S

Q: What are the options for this this exam Chemistry question?

A. 1 g · B. 2 g · C. 21 g · D. 42 g

Question

Q.51. The mass of iron converted into FeO by the action of 18 g of steam is : (Given : Molar mass of H, O and Fe are 1, 16 and 56 g mol respectively) Assume iron is present in excess :

FirstInTest · Accepted Answer

The balanced chemical equation for the reaction is: $3\mathrm{Fe(s)} + 4\mathrm{H_2O(g)} ightarrow \mathrm{Fe_3O_4(s)} + 4\mathrm{H_2(g)}$. Molar mass of H$_2$O = $2 	imes 1 + 16 = 18$ g/mol. Molar mass of Fe = 56 g/mol. Given 18 g of steam (H$_2$O). Number of moles of H$_2$O = $\frac{18 	ext{ g}}{18 	ext{ g/mol}} = 1$ mol. From the stoichiometry, 4 moles of H$_2$O react with 3 moles of Fe. So, 1 mole of H$_2$O will react with $\frac{3}{4}$ moles of Fe. Mass of Fe required = $\frac{3}{4} 	ext{ mol} 	imes 56 	ext{ g/mol} = 3 	imes 14 = 42$ g. Since iron is present in excess, 42 g of iron will be converted.

Answer

1 g

Answer

2 g

Answer

21 g

Answer

42 g

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