Chem~~help!

Question

10.0g of copper(II) carbonate is added to 50cm^3 of 1.00M sulphuric acid. At the end of the reaction ,1150cm^3 of a certain gas are collected at room and pressure.

a) Calculate the theoretical volume of the gas which would be liberated at room temperature and pressure.

b) Explain any difference between the theoretical volume and the volume of gas collected.

c) Calculate the theoretical mass of copper(II) sulphate crystals CuSo4．5H2O that can be obtained.

(Relative atomic masses:H=1.0, C=12.0, O=16.0, S=32.0, Cu=63.5; molar volume of any gas at room temperature and pressure =24.0dm^3 mol^-1)

Answer 1

a) Equation : CuCO3(s) + H2SO4(aq) --- > CuSO4(aq) + H2O(l) +CO2(g)
No. of mole of CuCO3 = 10/ 63.5+12+16*3 = 0.08097 mole
No. of mole of H2SO4 = Volume* Molarity = 0.05*1 = 0.05mole

From the equation, no. of mole of CuCO3 : H2SO4 = 1:1
so CuCO3 is in excess

No. of mole of H2SO4 : No. of mole of CO2 = 1:1
so no. of mole of CO2 = 0.05mole
Volume of CO2 = 0.05mole *24
=1.2 dm^3

b) It's because some CO2 gas is dissolved in the solution, so the theoretical volume is higher than the volume of gas collected.

c) From the equation of (a), the CuSO4．5H2O is the crystal of CuSO4 produced.
so no. of mole of H2SO4 : no. of mole of CuSO4 = 1:1
so no. of mole of CuSO4 = 0.05mole

Theoretical mass of CuSO4 ．5H2O = 0.05*(63.5+32+16*4+5*18)
= 249.5g

Hope this help

2007-10-28 21:35:28 補充：
(c) part should be 0.05*(63.5 32 16*4 5*18) = 0.05*249.5 = 12.475g

Answer 2

a)
Molar mass of CuCO3 = 63.5 + 12 + 16x3 = 123.5 g mol-1
Molar volume of a gas at room T and P = 24 dm3 = 24000 cm3

CuCO3 + H2SO4 → CuSO4 + H2O + CO2
Mole ratio CuCO3 : H2SO4 : CO2 = 1 : 1 : 1
No. of moles of CuCO3 used = 10/123.5 = 0.081 mol
No. of moles of H2SO4 used = 1 x (50/1000) = 0.05 mol
CuCO3 is in excess, and H2SO4 is the limiting reactant.
No. of moles of CO2 = No. of moles of H2SO4 = 0.05 mol
Theoretical volume of CO2 = 0.05 x 24000 = 1200 cm3

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b)
Theoretical volume - Actual volume = 1200 - 1150 = 50 cm3

The difference is mainly due to two reasons :
(1) Some CO2­ dissolves in the solution. (major reason)
(2) Some H2SO­4 is unreacted.

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c)
Molar mass of CuSO4•5H2O
= 63.5 + 32 + 16x4 + (1x2 + 16)x5
= 249.5 g mol-1

CuCO3 + H2SO4 → CuSO4 + H2O + CO2
CuSO4 + 5H2O → CuSO4•5H2O
CuCO3 is in excess, and H2SO4 is the limiting reactant.
Mole ratio H2SO4 : CuSO4•5H2O = 1 : 1
No. of moles of H2SO4 used = 0.05 mol
Theoretical no. of moles of CuSO4•5H2O formed = 0.05 mol
Theoretical mass of CuSO4•5H2O formed = 249.5 x 0.05 = 12.48 g