Question

1.12 L of CO2 at STP is bubbled through an NaOH solution. Reaction occuring:
2NaOH + CO2 → Na2CO3.
What is the mass of Na2CO3 formed?
(a) 10.6 g
(b) 5.3 g
(c) 1.12 g
(d) 2.24 g​

Answer 1

The correct question may be like:

If 1.12 liters of carbon dioxide gas is passed through a solution of sodium hydroxide at standard temperature and pressure, a reaction occurs which produces sodium carbonate. The equation for this reaction is 2NaOH + CO2 → Na2CO3. What is the mass of sodium carbonate that is formed as a result of this reaction? Choose from the following options: (a) 10.6 grams, (b) 5.3 grams, (c) 1.12 grams, or (d) 2.24 grams.

"The mass of sodium carbonate that is formed as a result of this reaction is (b) 5.3 grams.

The given equation is 2NaOH + CO2 → Na2CO3, which means that 2 moles of sodium hydroxide react with 1 mole of carbon dioxide to produce 1 mole of sodium carbonate.

First, we need to find the number of moles of carbon dioxide present in 1.12 liters of the gas at standard temperature and pressure (STP). At STP, 1 mole of any gas occupies a volume of 22.4 liters. Therefore, the number of moles of CO2 present in 1.12 liters can be calculated as:

n = V/22.4

n = 1.12/22.4

n = 0.05 moles

Since 1 mole of CO2 produces 1 mole of Na2CO3, the number of moles of Na2CO3 produced in the reaction is also 0.05 moles.

The molar mass of Na2CO3 can be calculated as:

Molar mass of Na2CO3 = 2 x atomic mass of Na + atomic mass of C + 3 x atomic mass of O

Molar mass of Na2CO3 = 2(23) + 12 + 3(16)

Molar mass of Na2CO3 = 106 g/mol

Therefore, the mass of Na2CO3 produced can be calculated as:

Mass of Na2CO3 = number of moles x molar mass

Mass of Na2CO3 = 0.05 x 106

Mass of Na2CO3 = 5.3 grams

Hence, the mass of sodium carbonate that is formed as a result of this reaction is 5.3 grams."

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Answer 2

Answer:

in this Q correct answer is

(b) 5.3g