# Question 45d18

Jun 1, 2015

Your reaction will produce 90.3 g of water.

$C a {\left(O H\right)}_{2} \stackrel{\textcolor{red}{\text{heat}}}{\to} C a O + {H}_{2} O$

Notice that you have a $1 : 1$ mole ratio between calcium hydroxide and water. This means that, regardless of how many moles of the former react, you'll always produce the exact same number of moles of water.

Use calcium hydroxide's molar mass to determine how many moles are present in 371 g

371cancel("g") * "1 mole"/(74.1cancel("g")) = "5.01 moles" $C a {\left(O H\right)}_{2}$

This means that you get

5.01cancel("moles"Ca(OH)_2) * ("1 mole"H_2O)/(1cancel("mole"Ca(OH)_2)) = "5.01 moles"# ${H}_{2} O$

Now use water's molar mass to determine how many grams you'd get

$5.01 \cancel{\text{moles") * "18.02 g"/(1cancel("mole")) = color(green)("90.3 g}}$