Question

2. The following reaction takes place at high temperatures.
Cr2O3 + 2 Al → 2 Cr + Al2O3

If 42.7 g Cr2O3 and 9.8 g Al are mixed,
a. what is the mass of Al2O3 formed?
b. which between Cr2O3 and Al is the limiting reactant? Excess?
c. how many grams of the excess will be the leftover?
d. if 15.2 grams of Al2O3 were formed, what is the percent yield?

Answer 1

Cr₂O₃ + 2 Al → 2 Cr + Al₂O₃

The question b) should be answered first because knowing the limiting reactant is necessary to determine the mass of Al₂O₃ formed, which is in the question a).

b)

n (Cr₂O₃) = 42.7g / 151.99g/mol = 0.281 mol

n (Al) = 9.8g / 26.98g/mol = 0.36 mol

According to the equation, n (Al) = 2n (Cr₂O₃). Actually 0.36/0.281 < 2, so Al is the limiting reactant. Cr₂O₃ is an excess reactant respectively.

a)

According to the equation, n(Al₂O₃) = 1/2 * n(Al). Therefore,

m(Al₂O₃) = 1/2 * (9.8g / 26.98g/mol) * 101.96g/mol = 18.5 g

c)

m_lefteover (Cr₂O₃) = 151.99g/mol * (42.7g / 151.99g/mol) - 1/2 * (9.8g / 26.98g/mol) = 15.1 g

d) % yield = (15.2g / 18.5g) * 100% = 82.2%