Question

When 4 gram of a mixture of Nahco3 and NaCl is heated 0.66 gram co2 gases evolved. determine the %composition of the original mixture

Answer 1

2 NaHCO3 = Na2CO3 + CO2 + H2O


Molar mass of NaHCO3 = 23+1 + 12 + 16X3 = 84 g/mol

2 moles of NaHCO3 = 2x84 = 168g


Molar mass of CO2 = 12 + 16x2 = 44g/mol


44g of CO2 is produced from 168g of NaHCO3

0.66g of CO2 will be produced from = (0.66x168)/44
                                                                = 2.52g


Percentage of NaHCO3 in the mixture = (2.52/4)x100 = 63%


Percentage of NaCl in the mixture = 100 – 63 = 37%

Answer 2

There is a sample of NaHCO3 and NaCl weigh 4 gm.

now , let the mass of NaHCO3 in the sample be x grams .

So,the mass of NaCl must be (4 - x )gm.

take two equations.

#1. 2 NaHCO3 -> Na2CO3 + H2O + CO2

mass of CO2 given = 0.66 grams

converting it into moles

= given mass / molar mass

= 0.66 gm / 44 gm/mole

= 0.03 /2 mole

now relating it to the #1 .

1 Mole of CO 2 is produced by 2 moles of NaHCO3 .

So,0.03/2 mole of will be produced by 0.03/2 × 2 = 0.03 moles .

here we got the moles of NaHCO3 .

Mass of NaHCO3 = 0.03 × (molar mass

of NaHCO3)

= 0.03 × 84 = 2.52gm

% composition of NaHCO3 is

= (2.52 / 4 )× 100 = 63%

now the % comp. of NaCl is 100 - 63 = 37 %

JAI HIND