how many grams of agbr, would be required to produce 75.0 g nabr? 2agbr + na2s2o3 → ag2s2o3 + 2nabr agbr…
how many grams of agbr, would be required to produce 75.0 g nabr? 2agbr + na2s2o3 → ag2s2o3 + 2nabr agbr : 187.77 g/mol nabr : 102.89 g/mol ? g agbr
Answer
Answer:
136.4 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 1:1. So, $n_{AgBr}=n_{NaBr} = 0.729\ mol$
Step3: Calculate mass of AgBr
$m_{AgBr}=n_{AgBr}\times M_{AgBr}=0.729\ mol\times187.77\ g/mol\approx136.4\ g$