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A. 2g mixture of Na2CO3 and NaHCO3 on heating to constant weight gave 224ml of CO2 at NTP. The % weights of NaHCO3 and Na2CO3 in the mixture are ? B. A hydrocarbon contains 85.8% of C. The weight of 5.6 L of the same gas at STP is found to be 14g. The molecular formula of the compound is ? C. 1.5g of an alkaline earth metal reacts with 0.35g of nitrogen to form its nitride. The metal is ?

A. 2g mixture of Na2CO3 and NaHCO3 on heating to constant weight gave 224ml of CO2 at NTP. The % weights of NaHCO3 and Na2CO3 in the mixture are ?
 
B. A hydrocarbon contains 85.8% of C. The weight of 5.6 L of the same gas at STP is  found to be 14g. The molecular formula of the compound is ?
 
C. 1.5g of an alkaline earth metal reacts with 0.35g of nitrogen to form its nitride. The metal is ?

Grade:12

1 Answers

Samyak Jain
333 Points
5 years ago
A. Let mass of Na2CO3 be x g, then mass of NaHCO3 is (2 – x) g in the mixture.
NaHCO3 decomposes into NaOH and liberate CO2 gas unlike Na2CO3.
Number of moles of Na2CO3 = x /106 mol and that of NaHCO3 = (2 – x) /84 mol.
1 mole of NaHCO3 gives 1 mole of CO2 and thus, (2 – x) /84 moles of NaHCO3 will give (2 – x) /84 moles of CO2.
No. of moles of CO2 = given volume / 22.4 L = 0.224 /22.4 = 0.01 mol.
\therefore (2 – x) /84 = 0.01  \Rightarrow  2 – x = 0.84 g  or  x = 1.16 g
i.e. Mass of Na2CO3 is 1.16 g and that of NaHCO3 is 0.84 g.
% weight of Na2CO3 = (1.16 g / 2 g) x 100 = 58%
% weight of NaHCO3 = (0.84 g / 2 g) x 100 = 42%
 
B.  %    Atomic ratio      Simplest atomic ratio   
C  85.8  85.8 /12 = 7.15     7.15 / 7.15 = 1         \therefore Empirical formula of hydrocarbon is CH2.
H  14.2  14.2 / 1  = 14.2     14.2 / 7.15 \approx 2        Empirical mass is 12x1 + 1x2 = 14 g/mol
Volume of gas at STP = 5.6 L  Thus, its no. of moles = 5.6 / 22.4 = ¼.
Weight of gas = 14 g = No. of moles x Molar mass = 1/4 mol x Molar mass
\therefore Molar mass of compound is 14 g / (1/4 mol) = 56 g/mol.
We know that  Molecular formula / Empirical formula = Molecular mass / Empirical mass
\therefore Molecular formula / CH2 = 56 / 14 = 4
i.e. Molecular formula of the hydrocarbon is C4H8.
 
C. Let us denote the alkaline earth metal by M and consider x as its molar mass. Its valency is 2 and the formula of its nitride is M3N2.
Reaction :           3 M           +           N2            \rightarrow           M3N2
Given weight :     1.5 g                    0.35 g
No. of moles      1.5 / x              0.35 / 28 = 1/80
Now by stoichiometry, 1 mole of N2 reacts with 3 moles of metal M. So, 1/80 mole of N2 will react with 3/80 mole of M.
\therefore 1.5 / x = 3 / 80  i.e.  x = (1.5)(80) / 3 = 40 g/mol
Molar mass of metal is 40 g/mol. Hence, the metal is calcium (Ca).

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