You must use 64.43 g H₂O.
Balanced chemical equation: H₂O + CO₂ → H₂CO₃
Moles of CO₂ = 157.35 g CO₂ × (1 mol CO₂/44.01 g CO₂) = 3.5753 mol CO₂
Moles of H₂O = 3.5753 mol CO₂ × (1 mol H₂O/1 mol CO₂) = 3.5753 mol Fe
Mass of H₂O = 3.5753 mol H₂O × (18.02 g H₂O /1 mol H₂O) = 64.43 g H₂O
Answer: The mass of water required is, 64.4 grams
Explanation: Given,
Mass of carbon dioxide = 157.35 g
Molar mass of carbon dioxide = 44.01 g/mole
Molar mass of water = 18.02 g/mole
First we have to calculate the moles of [tex]CO_2[/tex].
[tex]\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}=\frac{157.35g}{44.01g/mole}=3.58moles[/tex]
Now we have to calculate the moles of water.
The balanced chemical reaction is,
[tex]CO_2+H_2O\rightarrow H_2CO_3[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]CO_2[/tex] react with 1 mole of [tex]H_2O[/tex]
So, 3.58 moles of [tex]CO_2[/tex] react with 3.58 moles of [tex]H_2O[/tex]
Now we have to calculate the mass of water.
[tex]\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O[/tex]
[tex]\text{Mass of }H_2O=(3.58mole)\times (18.02g/mole)=64.4g[/tex]
Therefore, the mass of water required is, 64.4 grams
