Answer:To determine the amount of grams of aluminum needed to achieve 7300 kJ of heat released in the reaction, we need to use the given enthalpy of the reaction and the molar mass of aluminum. First, let's convert the enthalpy of the reaction from kJ/mol to kJ/g. We know that the molar mass of aluminum is approximately 26.98 g/mol.
-850.2 kJ/mol ÷ 26.98 g/mol = -31.5 kJ/g. Now we can calculate the grams of aluminum needed to release 7300 kJ of heat. We'll set up a proportion using the enthalpy of the reaction:
-31.5 kJ/g = 7300 kJ/x g
Cross-multiplying, we get:
-31.5x = 7300
Solving for x, we find:
x = 7300 ÷ -31.5
x ≈ -231.75 g
Since mass cannot be negative, we can disregard the negative sign. Therefore, approximately 231.75 grams of aluminum are needed to achieve 7300 kJ of heat released in the reaction.
Step-by-step explanation: