6. how many grams of silver bromide are produced when 206 grams of cobalt (iii) nitrate are produced…

Explanation:

Step1: Calculate molar mass of $Co(NO_3)_3$

The molar mass of $Co(NO_3)3$: $M{Co(NO_3)_3}=58.93+(14.01 + 16.00\times3)\times3=58.93+(14.01 + 48.00)\times3=58.93 + 62.01\times3=58.93+186.03 = 244.96\ g/mol$.

Step2: Calculate moles of $Co(NO_3)_3$

Given mass of $Co(NO_3)_3$ is $m = 206\ g$. The number of moles $n=\frac{m}{M}=\frac{206\ g}{244.96\ g/mol}\approx0.841\ mol$.

Step3: Determine mole - ratio from the balanced equation

From the balanced equation $CoBr_3 + 3Ag(NO_3)\rightarrow3AgBr+Co(NO_3)3$, the mole - ratio of $AgBr$ to $Co(NO_3)3$ is $n{AgBr}:n{Co(NO_3)_3}=3:1$.

Step4: Calculate moles of $AgBr$

If $n_{Co(NO_3)3}=0.841\ mol$, then $n{AgBr}=3\times n_{Co(NO_3)_3}=3\times0.841\ mol = 2.523\ mol$.

Step5: Calculate mass of $AgBr$

The molar mass of $AgBr$ is $M_{AgBr}=107.87+79.90 = 187.77\ g/mol$. The mass of $AgBr$, $m_{AgBr}=n_{AgBr}\times M_{AgBr}=2.523\ mol\times187.77\ g/mol\approx473\ g$.

Answer:

473 g