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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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Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?
s2008m [1.1K]

Answer:

65.08 g.

Explanation:

  • For the reaction, the balanced equation is:

<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

4 0
3 years ago
Convert 6.93 x 1024 atoms of carbon to moles of carbon.
Aleksandr [31]

Answer: 11.5 moles of carbon

Explanation:

Based on Avogadro's law:

1 mole of any substance has 6.02 x 10^23 atoms

So, 1 mole of carbon = 6.02 x 10^23 atoms

Z moles = 6.93 x 10^24 atoms

To get the value of Z, cross multiply:

(6.93 x 10^24 atoms x 1mole) = (6.02 x 10^23 atoms x Z moles)

6.93 x 10^24 = (6.02 x 10^23 x Z)

Z = (6.93 x 10^24) ➗ (6.02 x 10^23)

Z = 1.15 x 10

Z = 11.5 moles

Thus, there are 11.5 moles of carbon.

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Which of the following has the largest radius?<br><br><br> Answer:K
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Answer:

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