$$2Ca(s) + O_2(g) → 2CaO(s)$$
From the equation, we can see that 2 moles of calcium react with 1 mole of oxygen to produce 2 moles of calcium oxide.
First, we need to convert the given mass of calcium (36.5 grams) to moles. The molar mass of calcium is 40.08 g/mol.
$$36.5 \text{ g Ca} \times \frac{1 \text{ mol Ca}}{40.08 \text{ g Ca}} = 0.910 \text{ mol Ca}$$
Next, we need to determine the limiting reactant. To do this, we will compare the number of moles of calcium and oxygen available.
Assuming we have enough oxygen gas, we can calculate the maximum moles of calcium oxide that can be produced using the moles of calcium:
$$0.910 \text{ mol Ca} \times \frac{2 \text{ mol CaO}}{2 \text{ mol Ca}} = 0.910 \text{ mol CaO}$$
Therefore, 0.910 moles of calcium oxide will be produced when 36.5 grams of calcium reacts completely with oxygen gas.
