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RideAnS [48]
3 years ago
13

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

Chemistry
1 answer:
Sav [38]3 years ago
5 0

Answer:

1 mol Al; 1 mol O2

Explanation:ol Al; 10 mol O2

15.4 mol Al; 10.7 mol O2

or the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

4Al(s) + 302(g) —> 2Al2O3(s)

Express your answer aor the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

4Al(s) + 302(g) —> 2Al2O3(s)

Express your answer as a chemical formula.

1 mol Al; 1 mol O2

4 mol Al; 2.5 mol O2

12 mol Al; 10 mol O2

15.4 mol Al; 10.7 mol O2

Hold on, our servers are swamped. Wait for your answer to fully load.s a chemical formula.

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In order to <u>convert from moles of Al₂O₃ into moles of Al</u>, we'll need to use<em> the stoichiometric coefficients</em>, using a conversion factor that has Al₂O₃ moles in the denominator and Al moles in the numerator:

  • 0.400 mol Al₂O₃ * \frac{4molAl}{2molAl_2O_3} = 0.800 mol Al

So the correct answer is option C).

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