In a reaction: \(SnO_{2} + 2C → Sn + 2CO\) the mass of coke containing 80% carbon required to reduce 0.302 kg of pure tin oxide is?
(Sn - 119, O - 16, C - 12).
0.40 kg
0.20 kg
0.06 kg
0.04 kg
Explanation
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oohh...sory it is 0.06.myschool pls correct diz....the carbon required is 0.048 which is the 80% of 0.06g of coke....
The correct and undisputed explanation is:
For the reaction
SnO2+2C -> Sn+2CO
Since the question requires us to apply the mass of coke containing 80% carbon for the reduction of pure tin oxide.
Let's then find the mass of the coke containing 80% carbon or simply 80% of the molar mass of carbon
I.e.... percentage composition (80%)=M/12 *100 (if calculated) M becomes 9.6g of coke
Therefore, according to the equation of reaction
2moles of C=1mole of SnO2
2*9.6g=151g
19.2g=151g
Therefore, reason now, if 19.2g of coke (according to the equation of reaction) is the mass of coke containing 80% of carbon required to reduce the 1mole of pure tin oxide (151g) then how much of coke which will still contain 80% carbon that will then reduce 0.302kg of pure tin oxide
It then becomes
19.2g=151g
Xg=0.302g
Cross multiply
It becomes, 0.302*19.2/151 = 0.0384 ~0.04(D)
Here is an explanation:
Mass of coke is required not mass of carbon,so it is (0.048×100)÷80=0.06(c)
REF: Check:ababio chemistry
