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Multiple Choice
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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Verified step by step guidance
1
Start by writing the balanced chemical equation for the reaction between calcium and oxygen to form calcium oxide: Ca + O₂ → CaO.
Calculate the moles of calcium used in the reaction. Use the molar mass of calcium (approximately 40.08 g/mol) to convert grams to moles: moles of Ca = 112.6 g / 40.08 g/mol.
Calculate the moles of oxygen used in the reaction. Use the molar mass of oxygen (approximately 32.00 g/mol for O₂) to convert grams to moles: moles of O₂ = 24.0 g / 32.00 g/mol.
Determine the limiting reactant by comparing the mole ratio from the balanced equation to the moles calculated. The balanced equation shows a 1:1 ratio between Ca and O₂. Compare the moles of Ca and O₂ to find which is the limiting reactant.
Calculate the amount of oxygen gas remaining after the reaction. Subtract the moles of oxygen that reacted (based on the limiting reactant) from the initial moles of oxygen, then convert the remaining moles back to grams using the molar mass of oxygen.