General Chemistry

Question

Ca(OH)2 + 2HBr ---> CaBr + 2H2O

If 4 moles of Ca(OH)2 & moles of HBr was mixed

1) what is the limiting reagent

2) the number of moles CaBr2 produced

3) the grams of CaBr2 produced

Answer 1

First, we need to correct the formula for calcium bromide, which is CaBr₂, NOT CaBr (important to know so we can determine the correct molecular weight later).

Ca(OH)₂ + 2HBr --> CaBr₂ + 2H₂O

We also should know that the limiting reactant is the reactant that PRODUCES THE LEAST AMOUNT OF PRODUCT.

Solution:

Step 1: Calculate how many moles of Calcium Bromide (CaBr₂) are produced from Ca(OH)₂.

1 mole CaBr₂

4 moles Ca(OH)₂ x ----------------------- = 4 moles CaBr₂

1 mole Ca(OH)₂

Step 2: Calculate how many moles of CaBr₂ are produced from HBr.

1 mole CaBr₂

4 moles HBr x ---------------------- = 2 moles CaBr₂

2 moles HBr

Step 3: Determine the limiting reactant based on which reactant produced the least amount of CaBr₂.

The limiting reactant determines how much CaBr₂ is produced.

1. HBr is the limiting reactant since it produces the least amount of CaBr2.
2. Therefore, 2 moles of CaBr2 are produced.

Step 4: Calculate the mass (grams) of CaBr₂ by multiplying the number of moles by its molecular weight (molar mass).

199.89 g CaBr₂

2 moles CaBr₂ x ---------------------- = 399.78 g CaBr₂

1 mole CaBr₂