For the reaction 2HNO_3 + Mg(OH)_2 -> Mg(NO_3)_2 + 2H_2O2HNO3+Mg(OH)2→Mg(NO3)2+2H2O, how many grams of magnesium nitrate are produced from 8 moles of water?

1 Answer
Daniel M
Jun 2, 2018

593.2 " g "Mg(NO_3)_2593.2 g Mg(NO3)2

Explanation:

We can use stoichiometry to find the moles of Mg(NO_3)_2Mg(NO3)2 and then the respective grams of Mg(NO_3)_2Mg(NO3)2.

If 8 moles of water are produced, we can find out how many moles of Mg(NO_3)_2Mg(NO3)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 Mg(NO_3)_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 Mg(NO_3)_2 by referencing a periodic table and summing the masses of each element in the compound.)

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