Chemistry: How many moles of magnesium hydroxide are needed to produce 74.7 milliliters of water, if the density of water is 0.987 g/ml? show all steps of your calculations as well as the final answer. hno3 + mg(oh)2 → h2o + mg(no3)2?

.02047 = x / 0.02105 = 0.0004309mol...

How many moles of magnesium hydroxide are needed

DaePoggers

12.03.2022

Given volume of water produced in the reaction = 74.7 mL

Density of water = 0.987 g/mL

Calculating the mass of water = 74.7 mL * 0.987 g/mL = 73.73 g

Moles of water produced = $73.73 g H_{2}O * \frac{1 mol H_{2}O}{18.02 g H_{2}O} = 4.09 mol H_{2}O$

The balanced chemical equation between nitric acid and magnesium hydroxide is,

$2 HNO_{3}(aq) + Mg(OH)_{2}(aq) -- 2 H_{2}O(l) + Mg(NO_{3})_{2} (aq)$

1 mol Magnesium hydroxide gives 2 mol water.

Moles of Magnesium hydroxide = $4.09 mol H_{2}O * \frac{1 mol Mg(OH)_{2}}{2 mol H_{2}O} = 2.045 mol Mg(OH)_{2}$

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