QuestionMay 15, 2025

What mass of AgBr is formed when 35.5 mL of 0.184 M AgNO_(3) is treated with an excess of aqueous hydrobromic acid? 53.69 1.239 1.44 g 34.5g

What mass of AgBr is formed when 35.5 mL of 0.184 M AgNO_(3) is treated with an excess of aqueous hydrobromic acid? 53.69 1.239 1.44 g 34.5g
What mass of AgBr is formed when 35.5 mL of 0.184 M AgNO_(3) is treated with an excess of aqueous hydrobromic acid? 53.69 1.239 1.44 g 34.5g

Solution
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Answer

1.227 g Explanation 1. Calculate moles of AgNO_3 Use the formula: \text{moles} = \text{volume (L)} \times \text{molarity}. Convert 35.5 mL to liters: 35.5 \, \text{mL} = 0.0355 \, \text{L}. Moles of AgNO_3 = 0.0355 \, \text{L} \times 0.184 \, \text{M} = 0.006532 \, \text{mol}. 2. Determine moles of AgBr formed The reaction is AgNO_3 + HBr \rightarrow AgBr + HNO_3. Moles of AgNO_3 equals moles of AgBr since the ratio is 1:1. Thus, moles of AgBr = 0.006532 mol. 3. Calculate mass of AgBr Use the formula: \text{mass} = \text{moles} \times \text{molar mass}. Molar mass of AgBr = 107.87 \, (\text{Ag}) + 79.90 \, (\text{Br}) = 187.77 \, \text{g/mol}. Mass of AgBr = 0.006532 \, \text{mol} \times 187.77 \, \text{g/mol} = 1.227 \, \text{g}.

Explanation

1. Calculate moles of $AgNO_3$<br /> Use the formula: $\text{moles} = \text{volume (L)} \times \text{molarity}$. Convert 35.5 mL to liters: $35.5 \, \text{mL} = 0.0355 \, \text{L}$. Moles of $AgNO_3 = 0.0355 \, \text{L} \times 0.184 \, \text{M} = 0.006532 \, \text{mol}$.<br />2. Determine moles of AgBr formed<br /> The reaction is $AgNO_3 + HBr \rightarrow AgBr + HNO_3$. Moles of $AgNO_3$ equals moles of AgBr since the ratio is 1:1. Thus, moles of AgBr = 0.006532 mol.<br />3. Calculate mass of AgBr<br /> Use the formula: $\text{mass} = \text{moles} \times \text{molar mass}$. Molar mass of AgBr = $107.87 \, (\text{Ag}) + 79.90 \, (\text{Br}) = 187.77 \, \text{g/mol}$. Mass of AgBr = $0.006532 \, \text{mol} \times 187.77 \, \text{g/mol} = 1.227 \, \text{g}$.
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