in the reaction below, 1.000×10³ g lioh is combined with 8.80×10² g co₂. the reaction produces 3.25×10² g…

in the reaction below, 1.000×10³ g lioh is combined with 8.80×10² g co₂. the reaction produces 3.25×10² g h₂o in an experiment. co₂ + 2lioh → li₂co₃ + h₂o what is the limiting reactant in this reaction? o lioh o co₂ done

in the reaction below, 1.000×10³ g lioh is combined with 8.80×10² g co₂. the reaction produces 3.25×10² g h₂o in an experiment. co₂ + 2lioh → li₂co₃ + h₂o what is the limiting reactant in this reaction? o lioh o co₂ done

Answer

Explanation:

Step1: Calculate molar masses

The molar mass of $LiOH$: $M_{LiOH}=6.94 + 16.00+ 1.01=23.95\ g/mol$. The molar mass of $CO_2$: $M_{CO_2}=12.01 + 2\times16.00 = 44.01\ g/mol$.

Step2: Calculate moles of reactants

The moles of $LiOH$, $n_{LiOH}=\frac{m_{LiOH}}{M_{LiOH}}=\frac{1.000\times 10^{3}\ g}{23.95\ g/mol}\approx41.75\ mol$. The moles of $CO_2$, $n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}=\frac{8.80\times 10^{2}\ g}{44.01\ g/mol}\approx20.0\ mol$.

Step3: Determine the mole - ratio from the balanced equation

From the balanced equation $CO_2 + 2LiOH\rightarrow Li_2CO_3+H_2O$, the mole - ratio of $CO_2$ to $LiOH$ is $1:2$.

Step4: Calculate the moles of $LiOH$ needed if all $CO_2$ reacts

If all $n_{CO_2} = 20.0\ mol$ of $CO_2$ reacts, the moles of $LiOH$ required, $n_{LiOH - required}=2\times n_{CO_2}=2\times20.0\ mol = 40.0\ mol$. Since we have $41.75\ mol$ of $LiOH$ and we only need $40.0\ mol$ of $LiOH$ for all the $CO_2$ to react, $CO_2$ is the limiting reactant.

Answer:

$CO_2$