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Lunna [17]
3 years ago
11

If you start with 512 grams of aluminum and 1147 grams of copper chloride to make aluminum chloride and copper, what is the limi

ting reagent? 2Al + 3CuCl -> 2AlCl3 + 3Cu
Chemistry
1 answer:
Setler [38]3 years ago
5 0

Answer:

Copper (II) chloride.

Explanation:

Hello,

In this case, considering the described reaction which is also given as:

2Al + 3CuCl_2 \rightarrow 2AlCl_3 + 3Cu

For us to identify the limiting reactant we first compute the available moles of aluminium:

n_{Al}=512gAl*\frac{1molAl}{27gAl}=19.0molAl

Next, we compute the consumed moles of aluminium by the 1147 grams of copper (II) chloride by using their 2:3 molar ratio:

n_{Al}^{consumed}=1147gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2}*\frac{2molAl}{3molCuCl_2}  =5.69molAl

Thereby, we can infer aluminium is in excess since less moles are consumed than available whereas the copper (II) chloride is the limiting reactant.

Best regards.

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2 years ago
|
lilavasa [31]

1.66 M is the concentration of the chemist's working solution.

<h3>What is molarity?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

In this case, we have a solution of Zn(NO₃)₂.

The chemist wants to prepare a dilute solution of this reactant.

The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.

We want to know the concentration of this diluted solution.

As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:

n_1= n_2  (1)

and we also know that:

n = M x V_2

If we replace this expression in (1) we have:

M_1 x V_1= M_2 x V_2

Where 1, would be the stock solution and 2, the solution we want to prepare.

So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is M_2:

4.93 x 210 =  620 xM_2

M_2 = 1.66 M

This is the concentration of the solution prepared.

Learn more about molarity here:

brainly.com/question/19517011

#SPJ1

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