Reduction of 1 mol of Cr2O3 with aluminum releases 534 kJ of heat. How much heat will be released if Cr2O3

Reduction of 1 mol of Cr2O3 with aluminum releases 534 kJ of heat. How much heat will be released if Cr2O3 with a mass of 7.6 g is reduced?

Given:
m (Cr2O3) = 7.6 g,
Q = 534 kJ,
n (Cr2O3) = 1 mol,
Q2 -?
Decision:
Let’s find the amount of the substance Cr2O3.
n = m: M.
M (Cr2O3) = 152 g / mol.
n = 7.6 g: 152 g / mol = 0.05 mol.
1 mol Cr2O3 – 534 kJ,
0.05 mol – x kJ,
x = (0.05 mol × 534 kJ): 1 mol = 26.7 kJ.
Cr2O3 + 2Al = 2Cr + Al2O3 + 26.7 kJ.
Answer: 26.7 kJ.