Question

1. mole-to-mole (reaction of iron and sulfur)iron metal (fe) reacts with sulfur ($s_8$) to produce iron(iii) sulfide ($fe_2s_3$).- balanced equation: - how many moles of iron (fe) are required to react completely with 0.250 moles of sulfur ($s_8$)?- moles of fe: $\boldsymbol{__________________}$ mol8. mass-to-mass (synthesis of sulfur trioxide)sulfur dioxide ($so_2$) reacts with oxygen gas ($o_2$) to produce sulfur trioxide ($so_3$).- balanced equation: - if 75.0g of sulfur dioxide ($so_2$) reacts completely, what mass (in grams) of sulfur trioxide ($so_3$) is produced?- mass of $so_3$: $\boldsymbol{__________________}$ g9. mass-to-mole (combustion of propane)propane ($c_3h_8$) burns in oxygen ($o_2$) to produce carbon dioxide ($co_2$) and water ($h_2o$).- balanced equation: - if 44.0g of propane ($c_3h_8$) is burned, how many moles of water ($h_2o$) are produced?- moles of $h_2o$: $\boldsymbol{__________________}$ mol10. mass-to-mass (neutralization reaction)aqueous phosphoric acid ($h_3po_4$) reacts with aqueous potassium hydroxide (koh) to produce aqueous potassium phosphate ($k_3po_4$) and water ($h_2o$).- balanced equation: - if 20.0g of phosphoric acid ($h_3po_4$) reacts completely, what mass (in grams) of potassium phosphate ($k_3po_4$) is produced?- mass of $k_3po_4$: $\boldsymbol{__________________}$ g

Explanation:

Problem 7: Mole-to-Mole (Iron and Sulfur)

Step1: Balance the chemical equation

Balance Fe and S atoms:

$$16\text{Fe} + 3\text{S}_8 ightarrow 8\text{Fe}_2\text{S}_3$$

Step2: Find mole ratio of Fe to S₈

Mole ratio: $\frac{n(\text{Fe})}{n(\text{S}_8)} = \frac{16}{3}$

Step3: Calculate moles of Fe

Substitute $n(\text{S}_8)=0.250\text{ mol}$:
$$n(\text{Fe}) = 0.250\text{ mol} \times \frac{16}{3} = \frac{4}{3} \approx 1.33\text{ mol}$$

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Problem 8: Mass-to-Mass (Sulfur Trioxide)

Step1: Balance the chemical equation

Balance S and O atoms:

$$2\text{SO}_2 + \text{O}_2 ightarrow 2\text{SO}_3$$

Step2: Calculate molar mass of SO₂

Molar mass: $M(\text{SO}_2)=32.07 + 2\times16.00 = 64.07\text{ g/mol}$

Step3: Find moles of SO₂

$$n(\text{SO}_2) = \frac{75.0\text{ g}}{64.07\text{ g/mol}} \approx 1.171\text{ mol}$$

Step4: Use mole ratio for SO₃

Mole ratio $\frac{n(\text{SO}_3)}{n(\text{SO}_2)}=1$, so $n(\text{SO}_3)=1.171\text{ mol}$

Step5: Calculate molar mass of SO₃

Molar mass: $M(\text{SO}_3)=32.07 + 3\times16.00 = 80.07\text{ g/mol}$

Step6: Find mass of SO₃

$$m(\text{SO}_3)=1.171\text{ mol} \times 80.07\text{ g/mol} \approx 93.8\text{ g}$$

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Problem 9: Mass-to-Mole (Propane Combustion)

Step1: Balance the chemical equation

Balance C, H, O atoms:

$$\text{C}_3\text{H}_8 + 5\text{O}_2 ightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$$

Step2: Calculate molar mass of C₃H₈

Molar mass: $M(\text{C}_3\text{H}_8)=3\times12.01 + 8\times1.008 = 44.094\text{ g/mol}$

Step3: Find moles of C₃H₈

$$n(\text{C}_3\text{H}_8) = \frac{44.0\text{ g}}{44.094\text{ g/mol}} \approx 0.998\text{ mol}$$

Step4: Use mole ratio for H₂O

Mole ratio $\frac{n(\text{H}_2\text{O})}{n(\text{C}_3\text{H}_8)}=4$:
$$n(\text{H}_2\text{O})=0.998\text{ mol} \times 4 \approx 4.00\text{ mol}$$

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Problem 10: Mass-to-Mass (Neutralization)

Step1: Balance the chemical equation

Balance H, P, O, K atoms:

$$\text{H}_3\text{PO}_4 + 3\text{KOH} ightarrow \text{K}_3\text{PO}_4 + 3\text{H}_2\text{O}$$

Step2: Calculate molar mass of H₃PO₄

Molar mass: $M(\text{H}_3\text{PO}_4)=3\times1.008 + 30.97 + 4\times16.00 = 97.994\text{ g/mol}$

Step3: Find moles of H₃PO₄

$$n(\text{H}_3\text{PO}_4) = \frac{20.0\text{ g}}{97.994\text{ g/mol}} \approx 0.2041\text{ mol}$$

Step4: Use mole ratio for K₃PO₄

Mole ratio $\frac{n(\text{K}_3\text{PO}_4)}{n(\text{H}_3\text{PO}_4)}=1$, so $n(\text{K}_3\text{PO}_4)=0.2041\text{ mol}$

Step5: Calculate molar mass of K₃PO₄

Molar mass: $M(\text{K}_3\text{PO}_4)=3\times39.10 + 30.97 + 4\times16.00 = 212.27\text{ g/mol}$

Step6: Find mass of K₃PO₄

$$m(\text{K}_3\text{PO}_4)=0.2041\text{ mol} \times 212.27\text{ g/mol} \approx 43.3\text{ g}$$

Answer:

1. Problem 7:

• Balanced Equation: $16\text{Fe} + 3\text{S}_8

ightarrow 8\text{Fe}_2\text{S}_3$

• Moles of Fe: $1.33$ mol

1. Problem 8:

• Balanced Equation: $2\text{SO}_2 + \text{O}_2

ightarrow 2\text{SO}_3$

• Mass of $\text{SO}_3$: $93.8$ g

1. Problem 9:

• Balanced Equation: $\text{C}_3\text{H}_8 + 5\text{O}_2

ightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$

• Moles of $\text{H}_2\text{O}$: $4.00$ mol

1. Problem 10:

• Balanced Equation: $\text{H}_3\text{PO}_4 + 3\text{KOH}

ightarrow \text{K}_3\text{PO}_4 + 3\text{H}_2\text{O}$

• Mass of $\text{K}_3\text{PO}_4$: $43.3$ g