Question

A 3.7 g sample of copper(II) carbonate is added to 25 cm3 of 2.0 mol dm–3 hydrochloric acid. Which volume of gas is produced under room conditions? A 0.60 dm3 B 0.72 dm3 C 1.20 dm3 D 2.40 dm3

Answer 1

Answer:

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Answer 2

Answer:

Solution:- (A) 0.56

CuCO

3

+2HCl→CuCl

2

+CO

2

+H

2

O

Molecular weight of CuCO

3

=123.5g

No. of moles in 3.7g of CuCO

3

=

123.5

3.7

=0.03 mol

Also,

2moldm

−3

HCl, i.e.,

No. of moles HCl in 1dm

3

volume =2

No. of moles of HCl in 25cm

3

=

1000

25

×2=0.05 mol

Here HCl is limiting reagent.

Therefore, from the above reaction,

Volume of gas produced when 2 moles of HCl reacts =22.4dm

3

Volume of gas produced when 0.05 moles of HCl reacts =

2

22.4

×0.05=0.56dm

3