# Given the equation: LiOH + HBr --> LiBr + H2O. If you start with ten grams of lithium hydroxide, how many grams of lithium bromide will be produced?

Nov 12, 2015

Ten grams of LiOH will produce$\approx \text{40 g LiBr}$

#### Explanation:

$\text{LiOH(aq)"+"HBr(aq)}$$\rightarrow$$\text{LiBr(aq)"+"H"_2"O(l)}$

Molar Masses of LiOH and LiBr
$\text{LiOH} :$(1xx6.941 "g/mol Li")+(1xx15.999 "g/mol O")+(1xx1.00794 "g/mol H")="23.94794 g/mol LiOH"

$\text{LiBr} :$(1xx6.941 "g/mol Li")+(1xx79.904"g/mol Br")="86.845 g/mol LiBr"

To complete this problem, we are going to go from mass LiOH to moles LiOH to moles LiBr to mass LiBr.

Mass LiOH to Moles LiOH
Since you didn't give a numeric value for ten grams, it will be written as $10$, with only one significant figure. I will round the final answer to one significant figure.

10cancel("g LiOH")xx(1"mol LiOH")/(23.94794cancel("g LiOH"))="0.41757 mol LiOH"

Mole LiOH to Mole LiBr
Multiply the mol LiOH times the mole ratio between LiBr and LiOH in the balanced equation, so that LiBr is in the numerator and LiOH is in the denominator so it will cancel.

0.41757cancel("mol LiOH")xx(1"mol LiBr")/(1cancel("mol LiOH"))="0.41757 mol LiBr"

Mole LiBr to Mass LiBr
Multiply the mol LiBr times the molar mass of LiBr.

$0.41757 \text{mol LiBr"xx(86.845"g LiBr")/(1"mol LiBr")="36.26 g LiBr"="40 g LiBr}$ (rounded to one significant figure)