To solve this problem, we need to use the balanced chemical equation and the stoichiometry of the reaction:
CaBr2 + 2 KOH -> Ca(OH)2 + 2 KBr
From the equation, we can see that 1 mole of CaBr2 reacts with 2 moles of KOH to produce 1 mole of Ca(OH)2 and 2 moles of KBr.
To determine the mass of CaBr2 consumed, we need to first calculate the number of moles of Ca(OH)2 produced:
n(Ca(OH)2) = m(Ca(OH)2) / M(Ca(OH)2)
where:
m(Ca(OH)2) = 96 g (given)
M(Ca(OH)2) = 74.09 g/mol (molar mass of Ca(OH)2)
n(Ca(OH)2) = 96 g / 74.09 g/mol = 1.296 mol
According to the balanced equation, 1 mole of Ca(OH)2 is produced from 1 mole of CaBr2. Therefore, the number of moles of CaBr2 consumed is also 1.296 mol.
n(CaBr2) = 1.296 mol
Finally, we can calculate the mass of CaBr2 consumed:
m(CaBr2) = n(CaBr2) x M(CaBr2)
where:
M(CaBr2) = 199.89 g/mol (molar mass of CaBr2)
m(CaBr2) = 1.296 mol x 199.89 g/mol = 259 g
Therefore, 259 grams of CaBr2 is consumed when 96 grams of Ca(OH)2 is produced.