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soldi70 [24.7K]
3 years ago
7

I have a buddy who recycles electronics, and isolates metals from the connector pins electrical boards. He isolates gold, for ex

ample, and purifies it the best he can, then sells it along with his other scrap metal. This last go around he was able to isolate 3.00 g of Au with a process he claims results in a 80.0% yield. If he is correct, what was his theoretical yield
Chemistry
1 answer:
Kisachek [45]3 years ago
3 0

Answer:

Theoretical yield = 3.75g

Explanation:

Percent yield is defined as one hundred times the ratio between actual yield and theoretical yield. The expression is:

Percent yield = Actual Yield / Theoretical Yield * 100

In the problem, your actual yield was 3-00g.

Percent yield is 80.0%.

Solving for theoretical yield:

80% = 3.00g / Theoretical yield * 100

Theoretical yield = 3.00g / 80.0% * 100

<h3>Theoretical yield = 3.75g</h3>
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A 0.964 gram sample of a mixture of sodium formate and sodium chloride is analyzed by adding sulfuric acid. The equation for the
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Answer: 67.8 %.

Explanation:

Okay, let us delve right into the solution to the question;

The balanced chemical reaction is given by the equation (1) below;

2 HCOONa + H2SO4 ---------> 2 CO + 2 H2O + Na2SO4. ----------------------------------------------------------------------------(1).

From the balanced chemical reaction in equation (1) above we can see that; 2 moles of HCOONa reacts with one moles of tetraoxosulphate acid, H2SO4 to produce 2 moles of carbonmonoxide,CO; 2 moles of water, H2O and 1 mole of sodium tetraoxosulphate, Na2SO4.

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STEP ONE : find the carbon monoxide,CO pressure; P(CO).

Using the formula below;

P(t) = P(CO) + P(H2O). Hence;

P(CO) = P(t) - P(H2O). Note that P(H2O)= 19.8 torr.

==>P(CO)= 752 torr - 19.8 torr = 732.2 torr.

STEP TWO: calculate the number of moles of Carbonmonoxide,CO.

Using the formula below;

Number of moles= pressure(P) × volume(v) / gas constant(R) × temperature (T).

That is, n= PV/RT.

n= 732 torr × 0.242 Litres/ 62.4 × 295.15.

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=> 2 moles of HCOONa = 2 moles of CO/ 2 moles of CO = 1 mol( HCOONa/ CO).

Then, 9.62 × 10^-3 mol of CO × 1 mol( HCOONa/ CO).

==> 9.62 × 10^-3 mol HCOONa × molar masss of HCOONa(68 grams/mol)

= 0.654 grams.

Therefore, the percentage of sodium formate in the original mixture = 0.654 grams/ 0.964 gram × 100 = 67.8 %.

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