# Question #bb3e8

Jan 4, 2017

The reaction produces 24 g of water.

$\text{2LiOH" → "Li"_2"O" + "H"_2"O}$

Step 2. Calculate the moles of $\text{LiOH}$.

$\text{Moles of LiOH" = 63 color(red)(cancel(color(black)("g LiOH"))) × ("1 mol LiOH")/(23.95 color(red)(cancel(color(black)("g LiOH")))) = "2.63 mol LiOH}$

Step 3. Calculate the moles of $\text{H"_2"O}$

$\text{Moles of H"_2"O" = 2.63 color(red)(cancel(color(black)("mol LiOH"))) × ("1 mol H"_2"O")/(2 color(red)(cancel(color(black)("mol LiOH")))) = "1.32 mol H"_2"O}$

4. Calculate the mass of $\text{H"_2"O}$.

$\text{Mass of H"_2"O" = 1.32 color(red)(cancel(color(black)("mol H"_2"O"))) × ("18.02 g H"_2"O")/(1 color(red)(cancel(color(black)("mol H"_2"O")))) = "24 g H"_2"O}$