Math

QuestionIn the following chemical reaction, 2 mol of A will react with 1 mol of B to produce 1 mol of A2 B\mathrm{A}_{2} \mathrm{~B} without anything left over: 2 A+B A2 B2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{~A}_{2} \mathrm{~B}
But what if you're given 2.8 mol of A and 3.2 mol of BB ? 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: 2.8mot A×1 mol A2 B2mot A=1.4 mol A2 B3.2mot m×1 mol A2 B1mot B=3.2 mol A2 B\begin{array}{l} 2.8 \operatorname{mot} \mathrm{~A} \times \frac{1 \mathrm{~mol} \mathrm{~A}_{2} \mathrm{~B}}{2 \operatorname{mot} \mathrm{~A}}=1.4 \mathrm{~mol} \mathrm{~A}_{2} \mathrm{~B} \\ 3.2 \mathrm{mot} \mathrm{~m} \times \frac{1 \mathrm{~mol} \mathrm{~A}_{2} \mathrm{~B}}{1 \operatorname{mot} \mathrm{~B}}=3.2 \mathrm{~mol} \mathrm{~A}_{2} \mathrm{~B} \end{array}
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 A2 B\mathrm{A}_{2} \mathrm{~B} can be formed from the given amounts.
Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al( s)+3Cl2( g)2AlCl3( s)2 \mathrm{Al}(\mathrm{~s})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{AlCl}_{3}(\mathrm{~s})
You are given 14.0 g of aluminum and 19.0 g of chlorine gas.
Part A
If you had excess chlorine, how many moles of of aluminum chloride could be produced from 14.0 g of aluminum Express your answer to three significant figures and include the appropriate units.
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Studdy Solution
If we have excess chlorine, 0.519 mol\text{0.519 mol} of aluminum chloride can be produced from 14.0 g\text{14.0 g} of aluminum.

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