Ask Question
20 December, 10:29

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.

+5
Answers (1)
  1. 20 December, 10:45
    0
    We need 28.5 mL of Calcium solution

    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers