Question

How many milliliters of calcium, with a density of 1.55 g/mL, are needed to produce 85.8 grams of calcium fluoride in the single replacement reaction below.

Answer 1

We need 28.5 mL of Calcium solution

Step-by-step explanation:

Step 1: Data given

Density = 1.55 g/mL

Mass of calcium fluoride (CaF2) = 85.8 grams

Molar mass of CaF2 = 78.07 g/mol

Step 2: The balanced equation

Ca + 2HF → CaF2 + H2

Step 3: Calculate moles CaF2

Moles CaF2 = mass CaF2 / molar mass CaF2

Moles CaF2 = 85.8 grams / 78.07 g/mol

Moles = 1.10 moles

Step 4: Calculate moles of Ca

For 1 mol mol CaF2 we need 1 mol Ca^2+

For 1.10 moles CaF2, we need 1.10 moles Ca^2+

Step 5: Calculate mass of Ca^2+

Mass Ca^2+ = moles Ca^2+ / molar mass Ca^2+

Mass Ca^2+ = 1.10 moles * 40.08 g/mol

Mass Ca^2+ = 44.1 grams

Step 6: Calculate volume of Ca^2+

Volume Calcium = mass calcium / density

Volume calcium = 44.1 grams / 1.55 g/mL

Volume calcium = 28.45 mL ≈ 28.5 mL

We need 28.5 mL of Calcium solution