# Question 8e899

Jul 20, 2017

"3074 g Mg"("NO"_3)_2

#### Explanation:

The first thing to do here is to figure out how many moles of magnesium nitrate are produced by the reaction,

You know by looking at the balanced chemical equation

$2 {\text{HNO"_ (3(aq)) + "Mg"("OH")_ (2(s)) -> "Mg"("NO"_ 3)_ (2(aq)) + 2"H"_ 2"O}}_{\left(l\right)}$

that in order to produce $1$ mole of magnesium nitrate, the reaction must consume $2$ moles of nitric acid and $1$ moles of magnesium hydroxide.

Assuming that you have enough magnesium hydroxide to go around, you can say that the reaction will produce

41.45 color(red)(cancel(color(black)("moles HNO"_3))) * ("1 mole Mg"("NO"_ 3)_ 2)/(2color(red)(cancel(color(black)("moles HNO"_3)))) = "20.725 moles Mg"("NO"_3)_2#

In order to convert this to grams, use the molar mass of magnesium nitrate

$20.725 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Mg"("NO"_3)_2))) * "148.3 g"/(1color(red)(cancel(color(black)("mole Mg"("NO"_3)_2)))) = color(darkgreen)(ul(color(black)("3074 g}}}}$

The answer is rounded to four sig figs, the number of sig figs you have for the number of moles of nitric acid that take part in the reaction.