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Na2CO3 + 2HCI → 2NaCI + H2O + CO2. Using the above equation, what volume of ca...


Question

Na2CO3 + 2HCI → 2NaCI + H2O + CO2. Using the above equation, what volume of carbondioxide, measured at s.t.p is liberated when 53g of sodium carbonate is dissolved in hydrochloric acid?

(1 mole of gas occupies 22.4 dm3 at s.t.p)

(Na = 23, C = 12, O = 16)

Options

A)
44.8 dm3
B)
11.2 dm3
C)
100.1 dm3
D)
3.0 dm3
E)
22.4 dm3

The correct answer is B.

Explanation:

Na2CO3 + 2HCI → 2NaCI +H2O + CO2
53 gm Na2CO3 will therefore liberate (22.4)/(106) x (53)/(1) = 11.2 dm3

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