Predict the amount of oxygen gas that will remain after the reaction of 112.6 grams of calcium with 24.0 grams of oxygen.
A
8.5 g
B
11.0 g
C
18.0 g
D
24.0 g
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1
Write the balanced chemical equation for the reaction: 2Ca + O_2 → 2CaO.
Calculate the moles of calcium (Ca) using its molar mass (40.08 g/mol): moles of Ca = 112.6 g / 40.08 g/mol.
Calculate the moles of oxygen (O_2) using its molar mass (32.00 g/mol): moles of O_2 = 24.0 g / 32.00 g/mol.
Determine the limiting reactant by comparing the mole ratio from the balanced equation. The reaction requires 1 mole of O_2 for every 2 moles of Ca.
Calculate the amount of oxygen gas that remains unreacted by using the initial moles of O_2 and subtracting the moles of O_2 that reacted with the available moles of Ca.