Step-by-step explanation:
To calculate the number of moles of strontium bromide in 12.0 g of calcium bromide, we first need to use the molar mass of calcium bromide (CaBr₂) to convert grams to moles, and then use the mole ratio between calcium bromide and strontium bromide to find the number of moles of strontium bromide.
The molar mass of calcium bromide is:
CaBr₂: Ca = 1 x 40.08 g/mol = 40.08 g/mol
Br₂ = 2 x 79.90 g/mol = 159.80 g/mol
Total: 40.08 + 159.80 = 199.88 g/mol
Next, we can use the molar mass to convert the mass of calcium bromide to moles:
12.0 g CaBr₂ x (1 mol / 199.88 g) = 0.0600 mol CaBr₂
The balanced chemical equation for the reaction between calcium bromide and strontium nitrate is:
3 CaBr₂ + Sr(NO₃)₂ → 2 SrBr₂ + 3 Ca(NO₃)₂
From this equation, we can see that 3 moles of calcium bromide react with 2 moles of strontium bromide. This gives us a mole ratio of:
3 mol CaBr₂ / 2 mol SrBr₂
To find the number of moles of strontium bromide in 0.0600 mol of calcium bromide, we can use this mole ratio:
0.0600 mol CaBr₂ x (2 mol SrBr₂ / 3 mol CaBr₂) = 0.0400 mol SrBr₂
Therefore, there are 0.0400 moles of strontium bromide (SrBr₂) in 12.0 g of calcium bromide.