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stira [4]
3 years ago
14

Aluminum and oxygen react according to the following equation: 4Al + 3O2 -> 2Al2O3 In a certain experiment, 4.6g Al was react

ed with excess oxygen and 6.8g of product was obtained. What was the percent yield of the reaction?
Chemistry
1 answer:
stiv31 [10]3 years ago
8 0

Answer:

Percent yield: 78.2%

Explanation:

Based on the reaction:

4Al + 3O₂ → 2Al₂O₃

<em>4 moles of Al produce 2 moles of Al₂O₃</em>

<em />

To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:

(Actual yield (6.8g) / Theoretical yield) × 100

Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:

4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.

As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:

0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,

In grams (Molar mass Al₂O₃ = 101.96g/mol):

0.0852 moles of Al₂O₃ × (101.96g / mol) =

<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>

Thus, Percent yield is:

(6.8g / 8.7g) × 100 =

<h3>78.2% </h3>
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Answer:  The new volume is 53.3 ml

Explanation:

The combined gas equation is,

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P_1 = initial pressure of gas =  50.0 kPa

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V_1 = initial volume of gas = 80.0 ml

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T_2 = final temperature of gas = 320K

Now put all the given values in the above equation, we get:

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Answer:

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