Question

For the reaction HNO3 + Mg(OH)2→ Mg(NO3)2 + H2O, how many grams of magnesium nitrate are produced from 8.00 mol of nitric acid, HNO3? Balance the equation

Answer 1

Answer:

2HNO3 + Mg(OH)2 --> 2H2O + Mg(NO3)2

593.2 grams Mg(NO3)2

Explanation:

First, balance the equation

The first thing I notice is that there are 2 nitrates on the reactant side so I would put a 2 in front of HNO3. Now the problem is that there are 4 hydrogens on the reactants side, so add a 2 to H2O.

Next is stoichiometry

We are looking for Mg(NO3)2 so use the balanced equation to create a mole ratio and then convert to grams.

8 mol HNO3 * 1 mol Mg(NO3)2/2 mol HNO3 * 148.3 g Mg(NO3)2/1 mol Mg(NO3)2 = 593.2 grams Mg(NO3)2

Answer 2

593.2 grams of Magnesium Nitrate [$Mg(NO_3)_2$] are produced.  

Balancing the equation we get:

$2HNO_3 + Mg(OH)_2 = 2H_2O + Mg(NO_3)_2$

Here we can see that $2HNO_3$  produces $1Mg(NO_3)_2$

It implies that to make 1 mole of $Mg(NO_3)_2$ we need 2 moles of $2HNO_3$

Therefore in 8 moles of  $2HNO_3$ , 4 moles of $Mg(NO_3)_2$will be produces.

Now, 1 mole of $Mg(NO_3)_2$ = 148.3 g

       4 moles of $Mg(NO_3)_2$ = 4 × 148.3 g = 593.2 g of magnesium nitrate will be produced.

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