IIT JEE , AIEEE Forum - Chemistry , Inorganic Chemistry

anchitsaini

@ashwin4u4ever
i couldn't understand ur method cos i don't find any cause for neutralisation to be taking place

dev_22oct

2g of mixture of carbonate, bicarbonate and chloride of Na, on heating produce 56 mL of CO2 at NTP. 1.6g of the same mixture required 25mL of N HCl solution 4 neutralization . Calculate the %age of Na2CO3, NaHCO3 and NaCl in the mixture from the above given data

im sorry for delay!!!

suppose Na₂CO₃ = Xg NaHCO₃=Y g Na₂SO₄=2-(X+Y)g

on heating only NaHCO3 W'll decompose to give CO2 as follows

2NaHCO₃---------->Na₂CO₃+ H₂0 +CO₂

2(84) 22400ml at stp

y g NaHCO₃ wll give CO₂ = 22400y/168 ml at stp

actual CO₂ produced at stp = 56ml=22400y/168

dev_22oct

y=.42

1.6 g of the mixture requires 1MHCl =25 ml

therefore 2 g of the mixture w'll require 1M HCl=25 x 2/1.6=31.25 ml

=.O31mole of HCl

Na₂CO₃ + 2HCl---->2NaCl+H₂0+CO₂

106g 2moles

NaHCO₃ + HCl------>NaCl +H₂0 +C0₂

84g 1mole

x g Na₂CO₃ require HCl=2x/106 moles

.84 g NaHCO₃ require HCl =1 * .84/84

dev_22oct

=.01

2x/106 + .001=.031 now i think u can calculate further..

BALGANESH

hello my friend don't go for the soln posted by other people here is a simple method
equivalent wt = that wt which combines with 8 g of Oxygen =( 112000 ML of oxygen as 24000 ml = 16 g )

CO2 ( 44 g ) combines with 16 g of oxygen

so mass of CO2 combining with 8 g = M /2

dev_22oct

is it 16 or 32

in CO2 32 from oxygen and 12 from carbon which makes 44

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