If you had excess chlorine, how many moles of aluminum chloride could be produced from 27.0 g of aluminum?
In the following chemical reaction, 2 mol of A will react with 1 mol of B to produce 1 mol of A2B without anything leftover:
2A + B → A2B
But what if you're given 2.8 mol of A and 3.2 mol of B? The amount of product formed is limited by the reactant that runs out first, called the limiting reactant. To identify the limiting reactant, calculate the amount of product formed from each amount of reactant separately:
Notice that less product is formed with the given amount of reactant A. Thus, A is the limiting reactant, and a maximum of 1.4 mol of A2B can be formed from the given amounts
Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:
2 Al(s) + 3 Cl2 (g) → 2 AlCl3(s)
You are given 27.0 g of aluminum and 32.0 g of chlorine gas.
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