Question

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?

Answer 1

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 * (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 * (1 mol Mg(OH)_(2))/(2 mol H_(2)O) = 2.045 mol Mg(OH)_(2)`