kosh_ki
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Hi, can anyone explain why the answer is C for the q below here? In particular, how do you know quickly that CaCO3 is the limiting reactant here, and why wouldn't CaCO3 in powder (increased surface area) result in a quicker rate of reaction than when in chip form?
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And separately, for the below question for which the answer is D, why is it line 2 for Experiment Q rather than line 3, given that yield should have doubled? I do understand Experiment R but am confused about Q.

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Thanks so much.
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charco
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(Original post by kosh_ki)
Hi, can anyone explain why the answer is C for the q below here? In particular, how do you know quickly that CaCO3 is the limiting reactant here, and why wouldn't CaCO3 in powder (increased surface area) result in a quicker rate of reaction than when in chip form?
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Your logic is fine in terms of rate of reaction BUT you have to make 8.8g of carbon dioxide. This is 0.2 mol.

Hence, you need at least 0.2 mol of calcium carbonate, so D would not make enough (0.1 mol)
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charco
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(Original post by kosh_ki)
And separately, for the below question for which the answer is D, why is it line 2 for Experiment Q rather than line 3, given that yield should have doubled? I do understand Experiment R but am confused about Q.

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Thanks so much.
Once again you have been caught out by the fact that the limiting reagent changes between the two experiments
Mol of magnesium is constant at 1.2/24 = 0.05 mol

In experiment Q, even though there are 0.04 x 2 = 0.08 mol of sulfuric acid, the maximum hydrogen moles = 0.05, hence line #2

In experiment R, there is 0.04 x 0.5 = 0.02 mol of sulfuric acid and it is now the limiting reagent. Hence line #5

Quite a demanding question.
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