How many mL of 0.1 M HCl are required to react completely with 1 g mixture of `Na_2CO_3` and `NaHCO_3` containing equimolar amounts of both ?

Explanation

Step 1. To calculate the number of moles of the components in the mixture.
Suppose mass of `Na_2CO_3` present in the mixture = x g
Mass of `NaHCO_3` present in the mixture = (1 – x) g
Molar mass of `Na_2CO_3 = 2 xx 23 + 12 + 3 xx 16 = 106 g "mol"^(-1)`
Molar mass of `NaHCO_3 = 23 + 1 + 12 + 3 xx 16 = 84 g "mol"^(-1)`
Number of moles of `Na_2CO_3` in x g ` = (x)/(106)`
Number of moles of `NaHCO_3` in (1 – x) g= `(1 - x)/(84)`
As mixture contains equimolar amounts of the two,
`(x)/(106) = (1-x)/(84) " or " 106 - 106x = 84x " or " x = (106)/(106 + 84) = 106/190 g = 0.558 g `
Number of moles of `Na_2CO_3 = (0.558)/(106) = 0.00526`
Number of moles of `NaHCO_3 = (1-0.558)/(84) = (0.442)/(84) = 0.00526`
Step 2. To calculate the number of moles of HCl required.
`Na_2CO_3 + 2HCl to 2NaCl + H_2O + CO_2`
`NaHCO_3 + HCl to NaCl + H_2O + CO_2`
`NaHCO_3 + HCl to NaCl + H_2O + CO_2`
1 mole of `Na_2CO_3` requires = 2 moles of HCI
` therefore ` 0.00526 mole of `Na_2CO_3` requires = 0.00526 x 2 moles = 0.01052 mole
1 mole of `NaHCO_3` requires = 1 mole of HCI
` therefore ` 0.0526 mole of `NaHCO_3` requires = 0.00526 mole
` therefore ` Total HCl required = 0.01052 + 0.00526 moles = 0.01578 moles
Step 3. To calculate volume of 0.1 M HCl.
0.1 mole of 0.1 M HCl are present in 1000 mL of HCl.
0.01578 mole of 0.1 M HCl will be present in = `(1000)/(0.01) xx 0.01578 = 157.8 mL` of HCl