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
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3
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Answered by
1
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
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