CaBr + KOH – Ca(OH), + KBr (balance first) What mass, in grams, of CaBr is consumed when 48.00 g of Ca(OH), is produced?

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129.73 g of CaBr₂

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.

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