1.84g of a mixture contains caco3 and mgco3.On treating the mixture with dilute HCl .448ml of CO2 is obtained at STP. calculate the percentage composition of the mixture
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Answer:
Correct option is
A
50
Let CaCO
3
bexg.
Then, MgCO
3
is(18.4−x)g.
CaCO
3
→CaO+CO
2
1 mol 1 mol 1 mol
100g 56g 44g
xg
100
44x
g
MgCO
3
→MgO+CO
2
1 mol 1 mol 1 mol
84g 40g 44g
(18.4-x)g
84
44(18.4−x)
g
Total amount of CO
2
=
100
44x
+
84
44(18.4−x)
= 8.8 (given).
This gives x = 10g.
∴CaCO
3
=10g=0.1mol.
MgCO
3
=18.4−10=8.4g=0.1mol.
Total moles of CaCO
3
and MgCO
3
=0.2mol.
∴CaCO
3
=50mol%.
MgCO
3
=50mol%.
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