If 0.150 moles of calcium oxide reacts with carbon dioxide gas, how many grams of calcium carbonate can be produced?

15 g of CaCo₃

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CaO + CO₂ —> CaCo₃

From the balanced equation above,

1 mole of CaO reacted to produce 1 mole of CaCo₃.

Next, we shall determine the number of mole of CaCo₃ produced by the reaction of 0.15 mole of CaO. This can be obtained as follow:

From the balanced equation above,

1 mole of CaO reacted to produce 1 mole of CaCo₃.

Therefore, 0.15 mole of CaO will also react to produce 0.15 mole of CaCo₃.

Finally, we shall determine the mass of 0.15 mole of CaCo₃. This can be obtained as follow:

Mole of CaCo₃ = 0.15 mole

Molar mass of CaCo₃ = 40 + 12 + (16×3)

= 40 + 12 + 48

= 100 g/mol

Mass of CaCo₃ =?

Mole = mass / Molar mass

0.15 = Mass of CaCo₃ / 100

Cross multiply

Mass of CaCo₃ = 0.15 × 100

Mass of CaCo₃ = 15 g

Thus, 15 g of CaCo₃ were obtained from the reaction.