Question

Question #0bb3d

Answer 1

`"0.3 moles H"_2"O"`

Explanation:

Calcium hydroxide, `"Ca"("OH")_2`, will react with nitric acid, `"HNO"_3`, to form calcium nitrate, `"Ca"("NO"_3)_2`, a soluble salt that exists as ions in aqueous solution, and water.

In order to be able to say how many moles of water will result from this neutralization reaction, you need to write a balanced chemical equation first

`"Ca"("OH")_text(2(aq]) + 2"HNO"_text(3(aq]) -> "Ca"("NO"_3)_text(2(aq]) + color(red)(2)"H"_2"O"_text((l])`

Now, the problem doesn't provide you with information about how much acid you've got at your disposal, which means that you can assume that the nitric acid is in excess.

So, you have a `1:color(red)(2)` mole ratio between calcium hydroxide and water. This means that every mole of calcium hydroxide that takes part in the reaction will produce `color(red)(2)` moles of water.

Use calcium, hydroxide's molar mass to determine how many moles you have in that `"10-g"` sample

`10color(red)(cancel(color(black)("g"))) * ("1 mole Ca"("OH")_2)/(74.09color(red)(cancel(color(black)("g")))) = "0.1350 moles Ca"("OH")_2`

The reaction will produce

`0.1350color(red)(cancel(color(black)("moles Ca"("OH")_2))) * (color(red)(2)" moles H"_2"O")/(1color(red)(cancel(color(black)("mole Ca"("OH")_2)))) = "0.270 moles H"_2"O"`

Since you only provided one sig fig for the mass of calcium hydroxide, you can only use one sig fig for the answer

`n_(H_2O) = color(green)("0.3 moles H"_2"O")`