Question

THe volume in litres of CO2 gas at 298 K and a pressure of 765 mm Hg that would produce 0.3 kg of CaCO3 by the reaction with CA(oh)2?

Answer 1

Sorry can't help with chem. :/

Answer 2

same here sorry

Answer 3

is this from mole concept

Answer 4

no it is from states of matter!

Answer 5

\[try an use PV=nRT\]

Answer 6

mole concept :P

Answer 7

@A.Avinash_Goutham i think we need to use the temp and pressure to find no of moles

Answer 8

no like we got .3 kg r8...... so we get the moles.....the temp nd pressure are to indicate the stp i guess

Answer 9

@A.Avinash_Goutham yeah guessing is the best thing we can do right now

Answer 10

ca(oh)2 + c02 -->caco3 + h20

Answer 11

so 0.3 kg = 300 gms = 3 mole = 3*22.4?

Answer 12

ca(oh)2 + c02 -->caco3 + h20
12+ 32 = 44 g of co2 produces 40 + 12 + 48 = 100g of caco3

Answer 13

so by proportion
300g of caco3 is produced from 3*44 = 132g of co2

Answer 14

now we need to find what volume of co2 weighs 132 g at 298 degrees K and 765 mm of Hg

i can't remember what standard temperature and pressure is
1 mole of co2 (44g) has a volume of ( if my memory is correct) 22.4 litres at standard temp and pressure.
- sorry but i can't be definite about this
but 132 g of co2 is correct

Answer 15

ok - - just looked it up
standard temp and pressure is 760 mm hg and 273 degrees centigrade
and 22.4 liters is volume of 44g co2 at these values
so you need to use proportion to find required volume
132 g occupies 22.4 * 132/44 = 22.4 * 3 liters at STP
at 765mm and 298 degrees it is

22.4 * 3 * 298/273 * 760/765 liters

Answer 16

@ujjwal

Answer 17

@Kryten

Answer 18

hey sorry i came to see what's new and dont have time right now but in about 3 hrs ill be back and ill solve it step by step cause it is simple one...

Answer 19

ok

Answer 20

here you need 3 moles of \(CO_2\) to produce 0.3 kg \(CaCO_3\).
1 mole of \(CO_2\) at STP =22.4 liters.
Therefore, 3 mole of \(CO_2\) at STP= 3 \(\times\) 22.4 liters
Now. you have to find corresponding value of volume at 298 K and 765 mm Hg.

Use this relation:
\[\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\]

Put
\(P_1\)=760 mm Hg
\(V_1\)= 3 \(\times\) 22.4 liters
\(T_1\)=273 K

\(P_2\)=765 mm Hg
\(T_2\)=298 K
and find \(V_2\).
That will be your answer.