Ask Question
20 November, 01:05

Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45 g of NaHCO3 is present in an antacid tablet. Use stoichiometry (a mole ratio conversion must be present) to find your answers (there should be three: one answer for each balloon)

+3
Answers (1)
  1. 20 November, 04:16
    0
    For 1 antacid tablet (in ballon1) we get. 0173 moles of CO2

    for 2 tablets (in balloon 2) we get: 2*0,0173 = 0.0346 moles of CO2

    For 3 tablets (in balloon 3) we get 3 * 0.0173 = 0.0519 moles of CO2

    Explanation:

    The complete question:

    Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45 g of NaHCO3 is present in an antacid tablet. Use stoichiometry (a mole ratio conversion must be present) to find your answers (there should be three: one answer for each balloon).

    Balloon 1 had 1 antacid tab

    Baloon 2 had 2

    Balloon 3 had 3

    Step 1: Data given

    1.45 g of NaHCO3 is present in an antacid tablet

    Molar mass of NaHCO3 = 84.00 g/mol

    Step 2: The balanced equation

    NaHCO3 + H2O → NaOH + H2O + CO2

    Step 3: Calculate moles of NaHCO3

    Moles NaHCO3 = mass NaHCO3 / molar mass NaHCO3

    1.45g / 84.0 g/mol =.0173 moles

    Step 4: Calculate moles CO2

    For 1 mol NaHCO3 we need 1 mol H2O to produce 1 mol NaOH 1 mol H2O and 1 mol CO2

    For 0.0173 moles NaHCO3 we'll get 0.0173 moles CO2

    so for 1 antacid tablet we get. 0173 moles of CO2

    for 2 tablets we get: 2*0,0173 = 0.0346 moles of CO2

    For 3 tablets we get 3 * 0.0173 = 0.0519 moles of CO2
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45 g of ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers