Question

When 10.0 mol of calcium metal is reacted with 2.5 mol of oxygen gas, how much calcium oxide is produced? (Hint: write down the balanced equation before solving.)

Options:
a) 2.5 mol
b) 1.3 mol
c) 5.0 mol
d) -20.0 mol

Answer 1

When 10.0 mol of calcium reacts with 2.5 mol of oxygen, the balanced chemical equation shows the limiting reactant is oxygen. Since the reaction ratio is 1:1 for O₂ to CaO, 2.5 moles of oxygen gas produce 5.0 moles of calcium oxide.

Step-by-step explanation:

To determine how much calcium oxide is produced when 10.0 mol of calcium metal reacts with 2.5 mol of oxygen gas, we must first write down and balance the chemical equation. The balanced equation for the reaction between calcium metal and oxygen gas to form calcium oxide is:

2 Ca(s) + O2(g) → 2 CaO(s)

From the equation, we see that 2 moles of calcium metal react with 1 mole of oxygen gas to produce 2 moles of calcium oxide. Therefore, since the reaction ratio is 1:1, the limiting reactant will determine the amount of product formed.

In this case, the limiting reactant is oxygen since there are fewer moles of oxygen available (2.5 moles) compared to calcium (10.0 moles). Since 1 mole of oxygen yields 2 moles of calcium oxide, 2.5 moles of oxygen would yield:

2.5 moles O2 × 2 moles CaO / 1 mole O2 = 5.0 moles CaO

Therefore, the answer is (c) 5.0 moles of calcium oxide produced.