Question

14. An equirnolar mixture of Na2CO, and NaHCO, weighs 5g What is the mass of Na2CO3 in the mixture ? (RAM of Na=23, C= 12, O= 16,H= 1) O (a) 2.999 0 ( (b) 2.79g O (c)2.019 O (d) 2.219​

Answer 1

Formula: Na2CO3

Na Molecular Mass: 22.98 grams C Molecular Mass= 12.01 grams

O Molecular Mass= 16 grams Now, the molecular mass of the compound: 2(22.98) + 12.01 +3(16) = 105.97 grams

Now, when you do percent composition all you are doing is this: Amount of Element in Grams/ Total Compound Mass in Grams x 100%.

So, lets do that for each of these elements:

Na: 2 × 22.98 105.97 X 100% : 43.37%

12.01 105.97 C: x 100% = 11.33%

O: 3 x 16 105.97 x 100% = 45.30%

Now, add up these percentages and make sure that they add up to exactly 100%.

Answer 2

correct question-

An equirnolar mixture of Na2CO, and NaHCO, weighs 5g What is the mass of Na2CO3 in the mixture ?

$olution-

let the mass of Na2CO3 = x

so mass of NaHCO3= 5-x

1 mole of NaHCO3= 23+1+12+3×16

=84g

1 mole of Na2CO3= 2×23+12+3×16

=106g

Since they are equimolar

$\frac{5 - x}{x} = \frac{84}{106}$

$\frac{5 - x}{x} = \frac{42}{53}$

$265 - 53x = 42x$

$95x = 265$

$x = \frac{265}{95}$

x=2.79

(b) is correct.