A welding fuel gas contains carbon and hydrogen only . Burning a small sample of it in oxygen gives 3.38 g of carbondioxide , 0.69 g of water . A volume of 10 L of this welding gas is found to weigh 11.6 g calculate its
( i ) empirical formula
( i ) molecular formula
iii ) molar mass . 31 . Chlorine is prepared in the laboratory by the following method , 4HCI + MnO , ------- MnC , + CI + 2H2O How many grams of HCl is required to react completely with 250 ml of 0.25M Mno ,
Answers
Answer:
ANSWER
Since, 3.38 g of carbon dioxide are obtained, the mass of C present in the sample of welding fuel gas is
44
12
×3.38=0.92 g.
Since, 0.690 g of water are obtained, the mass of hydrogen present in the welding fuel gas sample is
18
2
×0.690=0.077g.
The percentage of C is
0.92+0.077
0.92
×100=92.3 %.
The percentage of H is
0.92+0.077
0.077
×100=7.7%.
(i) The number of moles of carbon =
12
92.2
=7.7.
The number of moles of hydrogen =
1
7.7
=7.7.
The mole ratio C:H=
7.7
7.7
=1:1.
Hence, the empirical formula of the welding fuel gas is CH.
(ii) The weight of 10.0 L of gas at N.T.P is 11.6 g.
22.4 L pf gas at N.T. P will weigh
10.0
11.6
×22.4=26g/mol
This is the molar mass.
(iii) Empirical formula mass is 12+1=13.
Molecular mass is 26.
The ratio, n of the molecular mass to empirical formula mass is
13
26
=2.
The molecular formula is 2(CH)=C
2
H
2
.
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