Question

Consider the reaction of N aHCO3 with an acid, HA. How many grams of NaHCO3 are required to produce 1.25 g of CO2 ?

Answer 1

Answer:

2.33g of NaHCO3

Explanation:

The balanced equation of the reaction is;

2NaHCO3  + 2HA --> 2CO2  +  2H2O + 2NaA

From the reaction above;

2 mol of NaHCO3 produces 2 mol of CO2

using mass = number of moles * molar mass

Mass = 2 * 84  = 164g of NaHCO3

Mass of CO2 = 2 * 44 = 88g of CO2

This means 164g of NaHCO3 produced 88g of CO2

How many grams of NaHCO3 would then produce 1.25g of CO2?

164 = 88

x = 1.25

upon solving for x;

x = 2.33g

Answer 2

The reaction of sodium bicarbonate with  acid produces equivalent moles of carbon dioxide. The mass of sodium bicarbonate in the given reaction is 2.352 grams.

What is stoichiometric law?

The stoichiometric law is used for the determination of the moles of each product or reactant, from the stoichiometric coefficient of the balanced chemical equation.

The balanced chemical equation for the reaction of sodium bicarbonate with HA acid is :

$\rm 2\;NaHCO_2\;+\;2\;HA\;\rightarrow\;2\;CO_2\;+\;2\;H_2O\;+\;2\;NaA$

From the stoichiometric law, 2 moles of carbon dioxide is produced from 2 moles of sodium bicarbonate.

The moles of carbon dioxide in 1.25 grams is:

$\rm Moles=\dfrac{mass}{molar\;mass}\\\\ Moles\;CO_2=\dfrac{1.25\;g}{44\;g/mol} \\\\Moles\;CO_2=0.028\;mol$

From the stoichiometric law, the moles of sodium bicarbonate required is:

$\rm 2\;mol\;CO_2=2\;mol\;NaHCO_3\\\\0.028\;mol\;CO_2=\dfrac{2\;mol}{2\;mol}\;\times\;0.028\;mol\;NaHCO_3\\\\ 0.028\;mol\;CO_2=0.028\;mol\;NaHCO_3$

The mass of sodium bicarbonate required is given as:

$\rm Mass=Moles\;\times\;Molar\;mass\\Mass\;NaHCO_3=0.028\;mol\;\times\;84\;g/mol\\Mass\;NaHCO_3=2.352\;g$

The mass of sodium bicarbonate in the reaction is 2.352 grams.

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