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Anonymous asked in Science & MathematicsChemistry · 4 weeks ago

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1. Using Stoichiometry determine how many moles of CO2 should be produced from 1 teaspoon of baking soda according to the nutritional label to find the initial amount of grams that you are starting with.   

A. 0.00714 mol of CO2

B. 0.057 mol of CO2

C. 403 mol of CO2

D. 0.000893 mol of CO2

2. Using Stoichiometry determine how many moles of CO2 should be produced from 2 teaspoon of baking soda according to the nutritional label to find the initial amount of grams that you are starting with. 

A. 0.114 moles of CO2

B. 0.00179 moles of CO2

C. 0.0143 moles of CO2

D. 806 moles of CO2

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  • 4 weeks ago

    Supposing the unnamed reaction to be something like:

    NaHCO3 + H{+} → CO2 + H2O + Na{+}  [balanced as written]

    (1 tsp) x (0.6 g / (1/8 tsp)) = 4.8 g = 5 g NaHCO3

    or

    (1 tsp) x ((150 x 10^-3 g Na) / (1/8 tsp)) / (22.98977 g Na/mol) x

    (1 mol NaHCO3 / 1 mol Na) x (84.0066 g NaHCO3/mol) = 4.38 g NaHCO3

    It's difficult to reconcile the two results above, although the difference indicates that there is something in the orange package which contains no sodium.  So for that reason, and because the second result has more significant digits, I will use 4.38 grams as the starting amount of NaHCO3:

    (4.38 g NaHCO3) / (84.0066 g NaHCO3/mol) x (1 mol CO2 / 1 mol NaHCO3) =

    0.0521 mol CO2

  • 4 weeks ago

    1 tsp = 4.8 g

    Molar mass NaHCO3 = 84 g/mol

    You don't indicate what the reaction is that is producing CO2. Is it the reaction with an acid? Or is it heating the NaHCO3 to decompose it? The reactions produce different results. I'm going to assume that it is the reaction with an acid. So,

    Reaction: NaHCO3 + H+ --> Na+ + CO2 + H2O

    4.8 g / 84 g/mol X (1 mol CO2 / 1 mol NaHCO3) = 0.057 mol CO2

    B. is the correct answer.

    If you double the mass of NaHCO3, you will double the moles of CO2 produced. So, 

    9.6 g NaHCO3 / 84 g/mol X (1 mol CO2/1 mol NaHCO3) = 0.114 mol CO2

    A is the correct answer.

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