5. Iron metal reacts with chlorine gas according to the following equation 2Fe(s

Question

5. Iron metal reacts with chlorine gas according to the following equation 2Fe(s) +3Cl(g)->2FeCl(s) Ir 25.6 g each of iron and chlorine are combined, how much FeCl should form? A. 74.3 g B. 113 g C. 39.0 g D. 49.5 g E. 26.0 g 5 6. What mass (in grams) of sSFe should be produced by the following reaction if 7.00 g of sulfur is mixed with 9.00 g of fluorine? A. 24.0 B. 6.40 D. 32.0 E. 16.0 Iron metal reacts with hydrochloric acid as follows 2 Fe(s) + 6HC?(w)-+2FeCh(ag) + 3H2(g) 7. f 35.6 g of iron react with excess HCI, and 98.6 g of FeCl, are collected, what is the percent yield of FeCI,? A. 103% B. 104% C. 95.7%

Explanation / Answer

5.

Mass of Fe = 25.6 g.

Molar mass of Fe = 56.0 g/mol

Moles of Fe = mass / molar mass = 25.6 / 56.0 = 0.457 mol

Mass of Cl2 = 25.6 g.

Molar mass of Cl2 = 71.0 g/mol

Moles of Cl2 = 25.6 / 71.0 = 0.360 mol

From the balanced equation,

2 mol of Fe needs 3 mol of Cl2

Then, 0.457 mol of Fe nneds 0.457 * 3 / 2 = 0.686 mol of Cl2 ( > 0.360 mol )

So, Cl2 is limiting reagent.

From the balanced equation,

3 mol of Cl2 forms 2 mol of FeCl3

Then, 0.360 mol of Cl2 forms 0.360 * 2 / 3 = 0.240 mol of FeCl3

Mass of FeCl3 = moles * molar mass = 0.240 * 162.5 = 39.0 g. of FeCl3 is formed.

So the answer is (C)

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