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Answer:
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.