Stoichiometry of an acid-base reaction Chemistry Lab 1. Suppose you started with

Question

Stoichiometry of an acid-base reaction Chemistry Lab

1. Suppose you started with 2.000 g of Na2CO3 (molar mass = 105.989 g/mol). How many moles of Na2CO3 were present in the beaker?

2. How many moles of HCl were added to the reaction every time 1.00 mL of solution was added? (n = M × V).

3. When you added the last amount of HCl that bubbled, how many moles of HCl had you added to the solution? Choose the closest answer. (7mL were added to the beaker)

4. Given the data below, how many grams of NaCl would you expect to be formed in the reaction of excess HCl with the Na2CO3? The molar mass of sodium carbonate is 105.989 g/mol and the molar mass of sodium chloride is 58.443 g/mol.

mass of empty beaker = 83.000grams mass of beaker plus NA2CO3 = 85.250grams mass of NA2CO3 = 2.250 grams

5. The molecular weight of NaCl is 58.44 g/mol. Based on the number of grams of NaCl recovered, how many moles of NaCl were generated? Choose the closest answer. (the number of grams recovered were 2.206 grams)

6. Based on the number of moles of Na2CO3 consumed and NaCl produced, what is the stoichiometry of the reaction?

7. Why did the reaction stop bubbling after adding 7.00 mL of HCl solution?

a. All of the NA2CO3 has been reacted and the additional HCL was just excess

b. All of the HCl was being converted to water

c. The HCl was in excess io the NA2CO3 under all conditions

d. The NA2CO3 was in excess to the HCl under all conditions

8. Based on the balanced chemical equation, given that 2.000 g of Na2CO3 (molar mass = 105.989 g/mol) were added to the reaction, how many grams of CO2 (molar mass = 44.01 g/mol) were produced? Remember that number of moles = mass (g) /molar mass (g/mol). Choose the closest answer.

9. Based on the balanced chemical equation, given that 2.000 g of Na2CO3 (molar mass = 105.989 g/mol) were added to the reaction, how many mL of H2O (molar mass = 18.02 g/mol) were produced? Remember that number of moles = mass (g) /molar mass (g/mol) and that the density of water is 1.000 g / mL. Choose the closest answer.

10. How does Na2CO3 act as a base?

a. The CO3 2- undergoes transmutation to OH-

b. The NA+ binds to H+ from water, forming OH- in solution

c. It acts as the conjugate acid

d. The CO3 2- binds to H+ from water, forming OH- in solution

Explanation / Answer

SOLUTION:

Note: Eight questions have been asked. Please break them in three sets. Here we attempt 5 - 6 questions.

(1) No. of moles of = given mass / molar mass = 2.0/105.98 = 0.018moles.

(2) No. of moles = Molarity X Volume. Molarity is not given here. Please insert the value of Molarity ypu forgot to write.

(3) Incomplete information is given.

(4) (i) Write balanced chemical equation.

Na2CO3 + 2HCl --------------> 2NaCl + H2O + CO2

(ii) Calculte moles of NaCl Produced.

One mole of Na2CO3 gives two moles of NaCl.

Or 105.98g of Na2CO3 give 2X58.44g (116.88g) of NaCl

therefore 2.250g of Na2CO3 will give = 116.88/105.98 X 2.250 = 2.48g of NaCl.

(5) Moles of NaCl Produced = Mass produced /Molar mass = 2.48/58.44 = 0.0424moles

(6) 0.021 moles of Na2CO3 are consumed and 0.042 moles of NaCl are produced. the ratio of moles of Vand NaCl is 1:2. Hence the stiochiometry will be same as shown above.

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