Question

A 2g sample containing na2co3 and nahco3 losses 0.248g when heated at 300°c,the temperature at which

Answer 1

At $300^{0}C$, NaHCO$_{3}$ decomposes to form gaseous carbon dioxide, water vapor and sodium carbonate.

The balanced chemical equation representing the decomposition of sodium bicarbonate is:

$2NaHCO_{3}(s) --> Na_{2}CO_{3}(s) + CO_{2}(g) + H_{2}O(g)$

Sodium bicarbonate is lost in the form of carbon dioxide gas and water vapor.

Mass of sodium bicarbonate lost = 0.248 g

Mass of $CO_{2} + H_{2}O = 0.248 g$

From the balanced chemical equation,

$2 mol NaHCO_{3} ---> 1 mol CO_{2} + 1 mol H_{2}O$

$2 mol * \frac{84 g NaHCO_{3}}{1 mol NaHCO_{3}}-> 1 mol CO_{2} * \frac{44 g CO_{2}}{1 mol CO_{2}} + 1 mol H_{2}O *\frac{18 g H_{2}O}{1 mol H_{2}O}$

$168 g NaHCO_{3} --> 62 g gaseous products (CO_{2} + H_{2}O)$

Mass of $NaHCO_{3}$ that would give 0.248 g $CO_{2} +H_{2}O$ = $0.248 g CO_{2} + H_{2}O * \frac{168 g NaHCO_{3}}{62 g CO_{2}+H_{2}O}$ = $0.672 g NaHCO_{3}$

Mass % of sodium bicarbonate in the sample = $\frac{0.672 g}{2g} * 100 = 33.6 % NaHCO_{3}$

Mass % of Sodium carbonate $Na_{2}CO_{3}$ = 100 - 33.6 % = 66.4 %

Answer 2

As reaction →  

                              2NaHCO3 → Na2CO3 + H2O + CO2

Moles                             2                   1             1        1

Molar Mass                    84                106         18       44

Mass                         2*84=16  1*106=106  1*18=18   1*44=44

As at 300 C 168g of NaHCO3 loosing the water & carbondioxide

     i.e, (18+44)=62g (total)

so, 0.48g can be lost by the =0.248*168 / 62 =0.672g of NaHCO3

Mass of Na2CO3 = Total mass of sample-Mass of NaHCO3

                            2 – 0.672 = 1.328g

% of Na2CO3 in the mixture = 1.328 / 2 * 100

                                               =66.4%