Question

What mass of water is required to react completely with 157.35 g CO2? (Molar mass of H2O = 18.02 g/mol; molar mass of CO2 = 44.01 g/mol)

Answer 1

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 2

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 $CO_2$.

$\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}=\frac{157.35g}{44.01g/mole}=3.58moles$

Now we have to calculate the moles of water.

The balanced chemical reaction is,

$CO_2+H_2O\rightarrow H_2CO_3$

From the balanced reaction we conclude that

As, 1 mole of $CO_2$ react with 1 mole of $H_2O$

So, 3.58 moles of $CO_2$ react with 3.58 moles of $H_2O$

Now we have to calculate the mass of water.

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=(3.58mole)\times (18.02g/mole)=64.4g$

Therefore, the mass of water required is, 64.4 grams