Question

The fizz produced when Alka Seltzer tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO₃) and citric acid (H₃C₆H₅O₇).

3NaHCO₃(aq) + H₃C₆H₅O₇(aq) → 3CO₂(g) + 3H₂O(l) + Na₃C₆H₅O₇(aq)

In a certain experiment 5.04 g of sodium bicarbonate and 4.75 g of citric acid are allowed to react.

a. How many grams of carbon dioxide form?

b. How many grams of the excess reactant remain after the limiting reactant is completely consumed?

Answer 1

a)

Molar mass of NaHCO3,

MM = 1*MM(Na) + 1*MM(H) + 1*MM(C) + 3*MM(O)

= 1*22.99 + 1*1.008 + 1*12.01 + 3*16.0

= 84.008 g/mol

mass(NaHCO3)= 5.04 g

use:

number of mol of NaHCO3,

n = mass of NaHCO3/molar mass of NaHCO3

=(5.04 g)/(84.01 g/mol)

= 5.999*10^-2 mol

Molar mass of C6H8O7,

MM = 6*MM(C) + 8*MM(H) + 7*MM(O)

= 6*12.01 + 8*1.008 + 7*16.0

= 192.124 g/mol

mass(C6H8O7)= 4.75 g

use:

number of mol of C6H8O7,

n = mass of C6H8O7/molar mass of C6H8O7

=(4.75 g)/(1.921*10^2 g/mol)

= 2.472*10^-2 mol

Balanced chemical equation is:

3NaHCO3(aq) + H3C6H5O7(aq) → 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq)

3 mol of NaHCO3 reacts with 1 mol of C6H8O7

for 5.999*10^-2 mol of NaHCO3, 2*10^-2 mol of C6H8O7 is required

But we have 2.472*10^-2 mol of C6H8O7

so, NaHCO3 is limiting reagent

we will use NaHCO3 in further calculation

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

According to balanced equation

mol of CO2 formed = (3/3)* moles of NaHCO3

= (3/3)*5.999*10^-2

= 5.999*10^-2 mol

use:

mass of CO2 = number of mol * molar mass

= 5.999*10^-2*44.01

= 2.64 g

Answer: 2.64 g

b)

According to balanced equation

mol of C6H8O7 reacted = (1/3)* moles of NaHCO3

= (1/3)*5.999*10^-2

= 2*10^-2 mol

mol of C6H8O7 remaining = mol initially present - mol reacted

mol of C6H8O7 remaining = 2.472*10^-2 - 2*10^-2

mol of C6H8O7 remaining = 4.726*10^-3 mol

Molar mass of C6H8O7,

MM = 6*MM(C) + 8*MM(H) + 7*MM(O)

= 6*12.01 + 8*1.008 + 7*16.0

= 192.124 g/mol

use:

mass of C6H8O7,

m = number of mol * molar mass

= 4.726*10^-3 mol * 1.921*10^2 g/mol

= 0.9079 g

Answer: 0.908 g