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.

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >