Question

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

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. a. 0.01024g b. 0.8305g c. 1.228g d. 8.305g

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. a. 0.2215mL b. 0.3400mL c. 0.1580mL d. 0.0191mL

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

Answer 1

Solution :-

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

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

Solution :- When Na2CO3 is reacting with HCl then it forms the CO2 gas which is bubbled out of the reaction mixture

Therefore when the Na2CO3 is completely reacted then added excess HCl do not show any reaction therefore it stops the formation of the bubbles.

So the correct answer is option ;a’

That is

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

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. a. 0.01024g b. 0.8305g c. 1.228g d. 8.305g

Solution :-

Mole ratio of the Na2CO3 to CO2 is 1 : 1

Therefore

(2.000 g Na2CO * 44.01 g CO2 / 105.989 g Na2CO3) = 0.8304 g CO2

So the correct answer is option b that is 0.8304 g CO2 can be formed

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. a. 0.2215mL b. 0.3400mL c. 0.1580mL d. 0.0191mL

Solution :-

Mole ratio of the Na2CO3 to H2O is 1 : 1

Therefore

(2.000 g Na2CO3 * 18.02 g H2O / 105.989 g Na2CO3 ) = 0.3400 g water

Now lets convert gram to ml of H2O

Volume = mass / density

              = 0.3400 g / 1.000 g per ml

              = 0.3400 ml

So the correct answer is option b. that is 0.3400 ml H2O

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

Solution:-

Reaction equation

CO3^2- + H2O    ----- > HCO3^- + OH-

Means CO3^2- abstract the proton from the water to form the HCO3^- and OH-

So the correct answer is option ‘d’ that is

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