Question

common polyatomic ions
chemical name chemical formula chemical name chemical formula
acetate $c_{2}h_{3}o_{2}^{-}$ nitrite $no_{2}^{-}$
carbonate $co_{3}^{2 - }$ ammonium $nh_{4}^{+}$
hypocarbonite $co^{2 - }$ cyanide $cn^{-}$
hydrogen carbonate (bicarbonate) $hco_{3}^{-}$ hydroxide $oh^{-}$
chlorite $clo_{2}^{-}$ peroxide $o_{2}^{2 - }$
hypochlorite $clo^{-}$ phosphate $po_{4}^{3 - }$
chlorate $clo_{3}^{-}$ hydrogen phosphate $hpo_{4}^{2 - }$
10
select the correct answer.
a solution with 117 grams of lithium hydroxide and another with 141 grams of hydrogen bromide are combined. they react according to this equation: $lioh + hbr
ightarrow libr + h_{2}o$.
what is the theoretical yield of lithium bromide? use the periodic table and polyatomic ion resource.
a. 117 g
b. 141 g
c. 151 g
d. 258 g
e. 424 g

Explanation:

Step1: Calculate moles of LiOH

The molar - mass of LiOH: $Li = 7\ g/mol$, $O = 16\ g/mol$, $H = 1\ g/mol$, so $M_{LiOH}=7 + 16+1=24\ g/mol$. Moles of $LiOH,n_{LiOH}=\frac{m}{M}=\frac{117\ g}{24\ g/mol}=4.875\ mol$.

Step2: Calculate moles of HBr

The molar - mass of HBr: $H = 1\ g/mol$, $Br = 80\ g/mol$, so $M_{HBr}=1 + 80 = 81\ g/mol$. Moles of $HBr,n_{HBr}=\frac{m}{M}=\frac{141\ g}{81\ g/mol}\approx1.74\ mol$.

Step3: Determine the limiting reactant

The balanced chemical equation is $LiOH + HBr
ightarrow LiBr + H_2O$. The mole - ratio of $LiOH$ to $HBr$ is $1:1$. Since $n_{HBr}<n_{LiOH}$, $HBr$ is the limiting reactant.

Step4: Calculate moles of LiBr

From the balanced equation, the mole - ratio of $HBr$ to $LiBr$ is $1:1$. So, moles of $LiBr,n_{LiBr}=n_{HBr}\approx1.74\ mol$.

Step5: Calculate the mass of LiBr

The molar - mass of $LiBr$: $Li = 7\ g/mol$, $Br = 80\ g/mol$, so $M_{LiBr}=7 + 80=87\ g/mol$. Mass of $LiBr,m = n\times M=1.74\ mol\times87\ g/mol\approx151\ g$.

Answer:

C. 151 g