Question

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What volume of CO2 gas at STP must be passed through 250ml 0.8M Na2CO3 solution to convert them into NaHCO3 solution?

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Answer 1

Answer:

$\large\boxed{\sf{4.48\;\;L}}$

Explanation:

First of all write the Reaction involved .

${Na}_{2}CO_{3} + CO_{2} + H_{2}O \rightarrow 2NaHCO_{3}$

Given that for ${Na}_{2}CO_{3}$

Volume = 250 ml

Molarity = 0.8 M

.°. Milimoles = 250 × 0.8 = 200

Now, in the reaction,

Let the Milimoles of $CO_{2}$ required is x.

Therefore, We get,

$= > 200 \times 3 + 2x + x = 6x \\ \\ = > 6x - 3x = 600 \\ \\ = > 3x = 600 \\ \\ = > x = 200$

Therefore, 200 Milimoles are required.

Now, at STP conditon,

1 mole = 22.4 L

$= > {10}^{ - 3} \: moles = 22.4 \times {10}^{ - 3} \: litres \\ \\ = > 1 \: milimole = 22 .4 \: \: ml \\ \\ = > 200 \: milimoles \: = 22.4 \times 200 \: ml = 4.48 \: litres$

Hence, 4.48 L is the reuired volume.

Answer 2

Explanation:4.48 L of CO2 gas may be required.

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