Ask Question
6 April, 07:33

A chemist adds 345.0 mL of a 0.520 mol/L of a barium chlorate solution to a reaction flask. Calculate the mass in grams of barium chlorate the chemist has added to the flask. Round your answer to significant digits.

+1
Answers (2)
  1. 6 April, 07:57
    0
    The mass of barium chlorate is 54.6 grams

    Explanation:

    Step 1: Data given

    Barium chlorate = Ba (ClO3) 2

    Volume of the barium chlorate solution = 345.0 mL = 0.345 L

    Concentration of the barium chlorate solution = 0.520 M

    Molar mass of barium chlorate = 304.23 g/mol

    Step 2: Calculate moles Ba (ClO3) 2

    Moles Ba (ClO3) 2 = volume of Ba (ClO3) 2 solution * concentration of Ba (ClO3) 2

    Moles Ba (ClO3) 2 = 0.345 L * 0.520 M

    Moles Ba (ClO3) 2 = 0.1794 moles

    Step 3: calculate mass of Ba (ClO3) 2

    Mass Ba (ClO3) 2 = moles Ba (ClO3) 2 * molar mass Ba (ClO3) 2

    Mass Ba (ClO3) 2 = 0.1794 moles * 304.23 g/mol

    Mass Ba (ClO3) 2 = 54.6 grams

    The mass of barium chlorate is 54.6 grams
  2. 6 April, 09:26
    0
    54.54g

    Explanation:

    First, let us obtain the number of mole of Barium chlorate, Ba (ClO3) 2 in the solution. This is illustrated below:

    Volume = 345mL = 345/1000 = 0.345L

    Molarity = 0.520mol/L

    Mole = ?

    Molarity = mole / Volume

    Mole = Molarity x Volume

    Mole = 0.520 x 0.345

    Mole = 0.1794mole

    Now, we can easily find the mass of the Barium chlorate. This is illustrated below:

    Molar Mass of Ba (ClO3) 2 = 137 + 2[35.5 + (16x3) ] = 137 + 2[35.5 + 48] = 137 + 2[83.5] = 137 + 167 = 304g/mol

    Number of mole of Ba (ClO3) 2 = 0.1794mole

    Mass = ?

    Mass = number of mole x Molar Mass

    Mass of Ba (ClO3) 2 = 0.1794 x 304

    Mass of Ba (ClO3) 2 = 54.54g
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A chemist adds 345.0 mL of a 0.520 mol/L of a barium chlorate solution to a reaction flask. Calculate the mass in grams of barium chlorate ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers