how many grams are in 1.11 moles of manganese sulfate, mn3(so4)7?

Explanation:

Step1: Calculate molar - mass of $Mn_3(SO_4)_7$

The molar - mass of $Mn$ is approximately $54.94\ g/mol$, the molar - mass of $S$ is approximately $32.07\ g/mol$, and the molar - mass of $O$ is approximately $16.00\ g/mol$. For $Mn_3(SO_4)_7$, there are 3 $Mn$ atoms, 7 $S$ atoms, and 28 $O$ atoms. $M = 3\times54.94+7\times32.07 + 28\times16.00$ $M=164.82+224.49+448$ $M = 837.31\ g/mol$

Step2: Calculate the mass

We use the formula $m = n\times M$, where $n = 1.11\ mol$ and $M = 837.31\ g/mol$. $m=1.11\ mol\times837.31\ g/mol$ $m = 929.4141\ g$

Answer:

$929.41\ g$ (rounded to two decimal places)