lioh + hbr → libr + h₂o\nif you start with ten grams of lithium hydroxide, how many grams of lithium bromide…

Explanation:

Step1: Calculate molar mass of LiOH

The molar mass of Li (lithium) is approximately $6.94\ g/mol$, O (oxygen) is $16.00\ g/mol$, and H (hydrogen) is $1.01\ g/mol$. So the molar mass of LiOH, $M_{LiOH}=6.94 + 16.00+1.01=23.95\ g/mol$.

Step2: Determine moles of LiOH

Given the mass of LiOH is $m = 10\ g$. Using the formula $n=\frac{m}{M}$, the number of moles of LiOH, $n_{LiOH}=\frac{10\ g}{23.95\ g/mol}\approx0.4175\ mol$.

Step3: Use mole - ratio from balanced equation

From the balanced chemical equation $LiOH + HBr\rightarrow LiBr + H_2O$, the mole - ratio of LiOH to LiBr is 1:1. So the number of moles of LiBr produced, $n_{LiBr}=n_{LiOH}\approx0.4175\ mol$.

Step4: Calculate molar mass of LiBr

The molar mass of Li is $6.94\ g/mol$ and Br (bromine) is $79.90\ g/mol$. So the molar mass of LiBr, $M_{LiBr}=6.94 + 79.90=86.84\ g/mol$.

Step5: Calculate mass of LiBr

Using the formula $m = n\times M$, the mass of LiBr, $m_{LiBr}=n_{LiBr}\times M_{LiBr}=0.4175\ mol\times86.84\ g/mol\approx36.2\ g$.

Answer:

$36.2\ g$