how many grams of agbr, would be required to produce 75.0 g nabr?\n2agbr + na₂s₂o₃ → ag₂s₂o₃ + 2nabr\nagbr…

how many grams of agbr, would be required to produce 75.0 g nabr?\n2agbr + na₂s₂o₃ → ag₂s₂o₃ + 2nabr\nagbr: 187.77 g/mol\nnabr: 102.89 g/mol\n? g agbr

how many grams of agbr, would be required to produce 75.0 g nabr?\n2agbr + na₂s₂o₃ → ag₂s₂o₃ + 2nabr\nagbr: 187.77 g/mol\nnabr: 102.89 g/mol\n? g agbr

Answer

Answer:

135.7 g

Explanation:

Step1: Calculate moles of NaBr

$n_{NaBr}=\frac{m_{NaBr}}{M_{NaBr}}=\frac{75.0\ g}{102.89\ g/mol}\approx0.729\ mol$

Step2: Determine mole - ratio from equation

From $2AgBr + Na_2S_2O_3\rightarrow Ag_2S_2O_3 + 2NaBr$, the mole - ratio of $AgBr$ to $NaBr$ is $\frac{n_{AgBr}}{n_{NaBr}}=\frac{2}{2} = 1$. So $n_{AgBr}=n_{NaBr}=0.729\ mol$

Step3: Calculate mass of AgBr

$m_{AgBr}=n_{AgBr}\times M_{AgBr}=0.729\ mol\times187.77\ g/mol\approx135.7\ g$