Question

question 32 (1 point)
if 100 g of h₂so₄ and 65 g of lioh are mixed, what is the limiting reactant?
h₂so₄ (aq) + 2 lioh (aq) → li₂so₄ (aq) + 2 h₂o (l)
lioh
h₂o
li₂so₄
h₂so₄
so₃

Explanation:

Step1: Calculate the molar mass of \(H_2SO_4\) and \(LiOH\)

The molar mass of \(H_2SO_4\): \(M_{H_2SO_4}=(2\times1)+32+(4\times16)=98\space g/mol\)
The molar mass of \(LiOH\): \(M_{LiOH}=7 + 16+1=24\space g/mol\)

Step2: Calculate the number of moles of \(H_2SO_4\) and \(LiOH\)

The number of moles of \(H_2SO_4\): \(n_{H_2SO_4}=\frac{m_{H_2SO_4}}{M_{H_2SO_4}}=\frac{100\space g}{98\space g/mol}\approx1.02\space mol\)
The number of moles of \(LiOH\): \(n_{LiOH}=\frac{m_{LiOH}}{M_{LiOH}}=\frac{65\space g}{24\space g/mol}\approx2.71\space mol\)

Step3: Use the stoichiometry of the reaction \(H_2SO_4(aq)+2LiOH(aq)\to Li_2SO_4(aq)+2H_2O(l)\)

From the balanced equation, the mole ratio of \(H_2SO_4\) to \(LiOH\) is \(1:2\).
For \(n_{H_2SO_4} = 1.02\space mol\), the moles of \(LiOH\) required \(n_{LiOH}^{required}=2\times n_{H_2SO_4}=2\times1.02 = 2.04\space mol\)
Since \(n_{LiOH}(2.71\space mol)>n_{LiOH}^{required}(2.04\space mol)\) and for \(n_{LiOH} = 2.71\space mol\), the moles of \(H_2SO_4\) required \(n_{H_2SO_4}^{required}=\frac{n_{LiOH}}{2}=\frac{2.71}{2}=1.36\space mol>1.02\space mol\)

Answer:

\(H_2SO_4\)