QuestionMay 2, 2025

5 C_(3)H_(8)+O_(2)arrow CO_(2)+H_(2)O If 20 gof propane burn in oxygen how many grams of water are produced? Chem (H) only: How many molecules of water are produced? a. Balance b. Given/Required c. Solution

5 C_(3)H_(8)+O_(2)arrow CO_(2)+H_(2)O If 20 gof propane burn in oxygen how many grams of water are produced? Chem (H) only: How many molecules of water are produced? a. Balance b. Given/Required c. Solution
5 C_(3)H_(8)+O_(2)arrow CO_(2)+H_(2)O If 20 gof propane burn in oxygen how many grams of water are produced? Chem (H) only: How many molecules of water are produced? a. Balance b. Given/Required c. Solution

Solution
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Answer

a. Balanced equation: C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O ### b. Required: Grams of water = 32.72 \, g, Molecules of water = 1.095 \times 10^{24} molecules Explanation 1. Balance the chemical equation The balanced equation is: C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O 2. Determine molar masses Molar mass of C_3H_8 = (3 \times 12) + (8 \times 1) = 44 \, g/mol Molar mass of H_2O = (2 \times 1) + 16 = 18 \, g/mol 3. Calculate moles of propane burned Moles of C_3H_8 = \frac{\text{mass}}{\text{molar mass}} = \frac{20}{44} \approx 0.4545 \, \text{mol} 4. Use stoichiometry to find moles of water produced From the balanced equation, 1 mole of C_3H_8 produces 4 moles of H_2O. Moles of H_2O = 0.4545 \times 4 = 1.818 \, \text{mol} 5. Convert moles of water to grams Mass of H_2O = \text{moles} \times \text{molar mass} = 1.818 \times 18 = 32.72 \, g 6. Calculate molecules of water Number of molecules = \text{moles} \times 6.022 \times 10^{23} = 1.818 \times 6.022 \times 10^{23} \approx 1.095 \times 10^{24} molecules.

Explanation

1. Balance the chemical equation<br /> The balanced equation is:<br />$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$<br /><br />2. Determine molar masses<br /> Molar mass of $C_3H_8 = (3 \times 12) + (8 \times 1) = 44 \, g/mol$<br /> Molar mass of $H_2O = (2 \times 1) + 16 = 18 \, g/mol$<br /><br />3. Calculate moles of propane burned<br /> Moles of $C_3H_8 = \frac{\text{mass}}{\text{molar mass}} = \frac{20}{44} \approx 0.4545 \, \text{mol}$<br /><br />4. Use stoichiometry to find moles of water produced<br /> From the balanced equation, 1 mole of $C_3H_8$ produces 4 moles of $H_2O$. <br />Moles of $H_2O = 0.4545 \times 4 = 1.818 \, \text{mol}$<br /><br />5. Convert moles of water to grams<br /> Mass of $H_2O = \text{moles} \times \text{molar mass} = 1.818 \times 18 = 32.72 \, g$<br /><br />6. Calculate molecules of water<br /> Number of molecules = $\text{moles} \times 6.022 \times 10^{23}$<br />$= 1.818 \times 6.022 \times 10^{23} \approx 1.095 \times 10^{24}$ molecules.
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