Answer:
129.73 g of CaBr₂
Step-by-step explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
CaBr₂ + 2KOH –> Ca(OH)₂ + 2KBr
Next, we shall determine the mass of CaBr₂ that reacted and the mass of Ca(OH)₂ produced from the balanced equation. This can be obtained as follow:
Molar mass of CaBr₂ = 40 + (80×2)
= 40 + 160
= 200 g/mol
Mass of CaBr₂ from the balanced equation = 1 × 200 = 200 g
Molar mass of Ca(OH)₂ = 40 + 2(16 + 1)
= 40 + 2(17)
= 40 + 34
= 74 g/mol
Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g
SUMMARY :
From the balanced equation above,
200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.
Finally, we shall determine the mass of CaBr₂ that react when 48 g of Ca(OH)₂ were produced. This can be obtained as follow:
From the balanced equation above,
200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.
Therefore, Xg of CaBr₂ will react to produce 48 g of Ca(OH)₂ i.e
Xg of CaBr₂ = (200 × 48)/74
Xg of CaBr₂ = 129.73 g
Thus, 129.73 g of CaBr₂ were consumed.