For the reaction 2HNO_3 + Mg(OH)_2 -> Mg(NO_3)_2 + 2H_2O, how many grams of magnesium nitrate are produced from 8 moles of water?

Jun 2, 2018

$593.2 \text{ g } M g {\left(N {O}_{3}\right)}_{2}$

Explanation:

We can use stoichiometry to find the moles of $M g {\left(N {O}_{3}\right)}_{2}$ and then the respective grams of $M g {\left(N {O}_{3}\right)}_{2}$.

If 8 moles of water are produced, we can find out how many moles of $M g {\left(N {O}_{3}\right)}_{2}$ are produced using this equation:

(8 " "cancel("mol "H_2O))/1*(1 " mol " Mg(NO_3)_2)/(2" "cancel("mol "H_2O))=4" mol "Mg(NO_3)_2

From here, we can use ratios to convert to the grams of $M g {\left(N {O}_{3}\right)}_{2}$:

(4 " "cancel("mol "Mg(NO_3)_2))/1*(148.3 " g " Mg(NO_3)_2)/(1 " "cancel("mol " Mg(NO_3)_2))=593.2 " g "Mg(NO_3)_2

(You can get the molar mass of $M g {\left(N {O}_{3}\right)}_{2}$ by referencing a periodic table and summing the masses of each element in the compound.)