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jekas [21]
3 years ago
13

What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?

Chemistry
1 answer:
AlladinOne [14]3 years ago
4 0

The question is incomplete, here is the complete question:

The given chemical reaction is:

2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3

What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?

<u>Answer:</u> The theoretical yield of copper is 44.48 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of aluminium = 12.6 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium}=\frac{12.6g}{27g/mol}=0.467mol

The given chemical equation follows:

2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of copper

So, 0.467 moles of aluminium will produce = \frac{3}{2}\times 0.467=0.7005mol of copper

Now, calculating the mass of copper  from equation 1, we get:

Molar mass of copper = 63.5 g/mol

Moles of copper = 0.7005 moles

Putting values in equation 1, we get:

0.7005mol=\frac{\text{Mass of copper}}{63.5g/mol}\\\\\text{Mass of copper}=(0.7005mol\times 63.5g/mol)=44.48g

Hence, the theoretical yield of copper is 44.48 grams

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<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
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<u>Step 1: Define</u>

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  1. Set up conversion:                     \displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \  H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})
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<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

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