1.33g impure CaC reacted with H2O. collected gas – ethyn. volume = 3.88 x 10-4 m3
calculated number of moles (at 284K, 101 kPa) correctly as 0.02 (by using ideal gas equation)
now
use previous answer to calculate the mass of calcium carbide (RMM – 64.1) when reacted with water. hence, calculate the % purity of the 1.33g sample of impure CaC.
CaC2 + 2H2O –> C2H2 + Ca(OH)2
Number of moles of C2H2 = 0.02
therefore moles of CaC2 reacted = 0.02 ( From balanced eqn mole ratio of CaC2 and C2H2 is 1 to 1)
Mass of CaC2 reacted = Moles X molar mass = .02 X 64.1=1.282g
% purity = (1.282 / 1.33) X 100 =96.4%
January 14th, 2010 - 7:59 pm
CaC2 + 2H2O –> C2H2 + Ca(OH)2
Number of moles of C2H2 = 0.02
therefore moles of CaC2 reacted = 0.02 ( From balanced eqn mole ratio of CaC2 and C2H2 is 1 to 1)
Mass of CaC2 reacted = Moles X molar mass = .02 X 64.1=1.282g
% purity = (1.282 / 1.33) X 100 =96.4%
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