Reaction Stoichiometry

Question

2Al + 3CuSO4 = Al2(SO4)3 + 3Cu

Moles of copper produced based on 0.633 g Al used

Answer 1

Let's assume that the limiting reagent is Al -- with the amount of Al (0.633g) you're given in the question. Because you're only given the amount of one of the reactant, this assumption is reasonable.

Also review the concept of limiting reagent - "The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. The limiting reagent is the one that is totally consumed; it limits the reaction from continuing because there is none left to react with the in-excess reactant."

Now convert 0.633g Al to moles of Al by dividing 0.633g by the molar mass of Al (which you can find in the periodic table).

Once you have the number of moles of Al used (let's you calculated this number as X), look at the coefficients of the reaction. For every 2 moles of Al consumed, 3 moles of Cu are produced.

Just do X * 3/2. That value is your final answer. Another way to think about it in terms of proportions: you can do X/Y = 2/3, where Y is the moles of copper produced. Then you get 2Y = 3X, and then Y = X * 3/2. Same as above.

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