Question

For the reaction NaBr (aq) + AgNO3 (aq) → AgBr (s) + NaNO3 (aq), how many liters of 4.5 M Silver Nitrate (AgNO3) would be required in order to mix with adequate Sodium Bromide (NaBr) to produce 250.0 g of Silver Bromide (AgBr)?

A. 0.25 mL NaNO3
B. 0.25 L NaNO3
C. 0.30 mL AgNO3
D. 0.30 L AgNO3
E. None of the Above

Answer 1

#75165 Chemistry, General Chemistry

For the reaction $\mathrm{NaBr}_{(\mathrm{aq})} + \mathrm{AgNO}_{3(\mathrm{aq})} \rightarrow \mathrm{AgBr}_{(\mathrm{s})} + \mathrm{NaNO}_{3(\mathrm{aq})}$ , how many liters of 4.5 M Silver Nitrate ( $\mathrm{AgNO}_3$ ) would be required in order to mix with adequate Sodium Bromide ( $\mathrm{NaBr}$ ) to produce 250.0 g of Silver Bromide ( $\mathrm{AgBr}$ )?

A. $0.25 \mathrm{~mL} \mathrm{NaNO}_{3}$

B. $0.25 \mathrm{~L} \mathrm{NaNO}_{3}$

C. $0.30 \mathrm{~mL} \mathrm{AgNO}_{3}$

D. $0.30 \mathrm{~L} \mathrm{AgNO}_{3}$

E. None of the Above

Answer:

$\mathrm{NaBr}_{(\mathrm{aq})} + \mathrm{AgNO}_{3(\mathrm{aq})} \rightarrow \mathrm{AgBr}_{(\mathrm{s})} + \mathrm{NaNO}_{3(\mathrm{aq})}$

$\mathrm{n} = \mathrm{m} / \mathrm{M}$

M (NaBr) = 102.9 g/mol

n (NaBr) = 250.0/102.9 = 2.4 mol

n $(\mathrm{AgNO}_3) = \mathrm{n}(\mathrm{NaBr}) = 2.4\mathrm{mol}$

$\mathrm{n} = \mathrm{C}_{\mathrm{M}}\cdot \mathrm{V}$ $\mathrm{V} = \mathrm{n} / \mathrm{C}_{\mathrm{M}}$

V $(\mathrm{AgNO}_3) = 2.4 / 4.5 = 0.51$

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