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3 years ago

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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 = 44.0

1 g/mol)

2 answers:

Ghella [55]3 years ago

6 0

You must use 64.43 g H₂O.

<em>Balanced chemical equation</em>: H₂O + CO₂ → H₂CO₃

<em>Moles of CO₂</em> = 157.35 g CO₂ × (1 mol CO₂/44.01 g CO₂) = 3.5753 mol CO₂

<em>Moles of H₂O</em> = 3.5753 mol CO₂ × (1 mol H₂O/1 mol CO₂) = 3.5753 mol Fe

<em>Mass of H₂O</em> = 3.5753 mol H₂O × (18.02 g H₂O /1 mol H₂O) = 64.43 g H₂O

zloy xaker [14]3 years ago

3 0

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

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Equation of the reaction:

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From the equation above, 1 mole of Cu reacts with 2 moles of AgNO3 to produce 2 moles of Ag and 1 mole of Cu(NO3)2.

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Explanation:

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$1 km=10^9 \mu m$

$175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m$

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