how many grams of ag2s2o3 form when 125.0 g agbr reacts completely according to the reaction below? 2agbr +…

Answer:

110.4 g

Explanation:

Step1: Calculate moles of AgBr

$n_{AgBr}=\frac{m_{AgBr}}{M_{AgBr}}=\frac{125.0\ g}{187.70\ g/mol}\approx0.666\ mol$

Step2: Determine mole - ratio

From the balanced equation $2AgBr + Na_2S_2O_3\rightarrow Ag_2S_2O_3 + 2NaBr$, the mole - ratio of $AgBr$ to $Ag_2S_2O_3$ is $2:1$. So, $n_{Ag_2S_2O_3}=\frac{1}{2}n_{AgBr}$. $n_{Ag_2S_2O_3}=\frac{1}{2}\times0.666\ mol = 0.333\ mol$

Step3: Calculate mass of $Ag_2S_2O_3$

$m_{Ag_2S_2O_3}=n_{Ag_2S_2O_3}\times M_{Ag_2S_2O_3}=0.333\ mol\times327.74\ g/mol\approx110.4\ g$