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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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When 6.85×10⁵ cal is converted to kilojoules, the result obtained is 2866.04 KJ

<h3>Data obtained from the question </h3>
  • Energy (cal) = 6.85×10⁵ cal
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<h3>Conversion scale </h3>

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<h3>How to convert 6.85×10⁵ cal to kilojoules</h3>

1 cal = 0.004184 KJ

Therefore,

6.85×10⁵ cal = 6.85×10⁵ × 0.004184

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