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uranmaximum [27]
3 years ago
9

The limiting reactant is the chemical substance that determines the amount of product(s) that can ultimately be formed in a reac

tion. During the reaction, the limiting reactant is completely consumed or used up, and therefore, causes the reaction to stop. The limiting reactant can be identified through stoichiometric calculations. After comparing the results, the reactant that produces the smaller mass of product is identified as the limiting reactant. Determine the limiting reactant and calculate the number of grams of Al2O3 that can be formed when 10.0 grams of Al and 5.00 grams of O2 react. 4 Al(s) + 3 O2(g) → 2 Al2O3(s) For the calculations in this module, the molar mass of an element will be rounded to the hundredths place (0.01 g).
Chemistry
1 answer:
Mariulka [41]3 years ago
6 0

Explanation:

Al=10,o2=5

(Al10)2(o5)3

The answer is oxygen.

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7 0
3 years ago
Twenty five grams of Iron 3 oxide react with an excess of carbon monoxide to form 15 g of Fe. Carbon dioxide is the other produc
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<h3>Answer:</h3>

Theoretical mass = 17.42 g

Percent yield of Fe = 86.11%

<h3>Explanation:</h3>

The equation for the reaction between iron (iii) oxide and carbon monoxide is given by;

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

We are required to calculate the theoretical yield and the percentage yield of Iron.

Step 1: Moles of iron (iii) oxide

Moles are given by dividing the mass of the compound by the molar mass.

Molar mass of Iron(iii) oxide = 159.69 g/mol

Moles of Iron(III) oxide = 25 g ÷ 159.69 g/mol

                                     = 0.156 moles

Step 2: Moles of Iron produced

From the equation 1 mole of Iron(iii) oxide reacts to produce 2 moles of Fe.

Therefore, the mole ratio of Fe₂O₃ to Fe is 1 : 2.

Thus, moles of Fe = Moles of Fe₂O₃ × 2

                              = 0.156 moles × 2

                              = 0.312 moles

Step 3: Theoretical mass of iron produced

To calculate the mass of iron we multiply the number of moles of iron with the relative atomic mass.

Relative atomic mass = 55.845

Mass of iron = 0.312 moles × 55.845

                    = 17.42 g

Step 4: Percent yield of iron

% yield = (Actual mass ÷ Theoretical mass)×100

            = (15 g ÷ 17.42 g) × 100 %

            = 86.11%

7 0
3 years ago
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