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# CIE AS level Chemistry watch

1. A 0.216 g sample of an aluminium compound X reacts with an excess of water to produce a single hydrocarbon gas. This gas burns completely in O2 to form H2O and CO2 only. The volume of CO2 at room temperature and pressure is 108cm3 . What is the formula of X?

A. Al2C3 B. Al3C2 C. Al3C4 D. Al4C3

Please help me with this question. Answer scheme says it's D.
2. What have you tried, so far?
3. (Original post by Pigster)
What have you tried, so far?
I tried using the general combustion equation but can't seem to get it. Can you please show me the working?
4. How many mol of C atoms are there in the 0.216 g of X?
5. A 0.216 g sample of an aluminium compound X reacts with an excess of water to produce a single hydrocarbon gas.
This gas burns completely in O2 to form H2O and CO2 only. The volume of CO2 at room temperature and pressure is 108cm3.
What is the formula of X?
A Al 2C3
B Al 3C2
C Al 3C4
D Al 4C3

Solution:

Volume of CO2 = 108cm3 = 108 ÷1000 dm3 = 0.108 dm3
No. of moles of CO2 in 108cm3 = 0.108 ÷ 24 = 0.0045 mol
No. of molecules of CO2 in 108cm3 = 0.0045 × 6.02×10^(23) = 2.709×10^(21)
As there is one carbon atom in every molecule of CO2:
No. of atoms of carbon in 108cm3 of C02= 2.709×10^(21)

No. of atoms of carbon must be same before and after the reaction.

A)
Mr of Al 2C3 = 2(27) 3(12) = 90
No. of moles in 0.216 g of Al 2C3 = 0.216 ÷ 90 = 0.0024 mol
No. of molecules in 0.216 g of Al 2C3 = 0.0024 × 6.02×10^(23) = 1.4448×10^(21)
As in every molecule of Al2C3, there are 3 Carbon atoms:
No. of atoms of carbon in 0.216 g of Al 2C3 = 4.3344×10^(21)
B)
Mr of Al 3C2 = 3(27) 2(12) = 105
No. of moles in 0.216 g of Al 3C2 = 0.216 ÷ 105 = 0.0020571429 mol
No. of molecules in 0.216 g of Al 3C2 = 0.0020571429 × 6.02×10^(23) = 1.2384×10^(21)
As in every molecule of Al3C2, there are 2 Carbon atoms:
No. of atoms of carbon in 0.216 g of Al 3C2 =1.2384×10^(21)×2 = 2.4768×10^(21)
C)
Mr of Al 3C4 =3(27) 4(12) = 129
No. of moles in 0.216 g of Al 3C4 = 0.216 ÷ 129 = 0.0016744186 mol
No. of molecules in 0.216 g of Al 3C4 = 0.0016744186 × 6.02×10^(23) = 1.008×10^(21)
As in every molecule of Al3C4, there are 4 Carbon atoms:
No. of atoms of carbon in 0.216 g of Al 3C4 = 1.008×10^(21) × 4 = 4.03199999×10^(21)
*D) -The correct Answer
Mr of Al 4C3 = 4(27) 3(12) = 144
No. of moles in 0.216 g of Al 4C3 = 0.216 ÷ 144 = 0.0015 mol
No. of molecules in 0.216 g of Al 4C3 = 0.0015 × 6.02×10^(23) = 9.03×10^(20)
As in every molecule of Al4C3, there are 3 Carbon atoms:
No. of atoms of carbon in 0.216 g of Al4C3 = 9.03×10^(20) × 3 = 2.709×10^(21)
6. OR just since C is -4 And Al is +3 just choose D
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