Na CO2 + 2HCI 2NaCl + CO, +H2O (a) moles of NaCl formed when 10.6 gm of Na, Co, is mixed with 100 ml of 0.5 M HCl solution. (b) Calculate the concentration of each ion in the solution after the reaction (C) Volume of CO, liberated at S.T.P.
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Answer:
A
HCl is in excess.
B
117.0 g of NaCl is formed.
C
The volume of CO
2
produced at 1 bar and 273 K is 22.4 L.
D
The volume of CO
2
produced at 1 bar and 298 K is 22.4 L.
Molecular weight of Na
2
CO
3
=106 g/mol, HCl=36.5g/mol and of NaCl =58.5 g/mol.
Moles of Na
2
CO
3
=
106
106
=1.0 mol and moles of HCl=
36.5
109.5
=3.0 mol.
A. Since, for 1 mol of Na
2
CO
3
, 2 mol of HCl is required.
So, HCl is in excess =(3−2)=1.0 mol
Therefore, Na
2
CO
3
is the limiting quantity.
B. Weight of NaCl formed =1×2×58.5=117.0 g NaCl
C. 1 mol of Na
2
CO
3
=1 mol of CO
2
=22.4L at 1 bar, 273 K.
D. 1 mol of Na
2
CO
3
=1 mol of CO
2
=22.4 L at 1 bar, 298 K.
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