Combining the two proportions in one:—
300: 273 ):: 1450: x = 1316cc. 760: 758 )
1316cc=1.316 liters. It remains to find the weight of this gas. A liter of H weighs 0.0896g. The vapor density of O is 16. Hence 1.316 liters of O will weigh 1.316 X 16 X 0.0896 =1.89g.
(KClO3 = KCl + O3)
From the equation (122.5 48) we make a proportion,
( 5 x)
122.5: 5:: 48: x = 1.95, and obtain, as the weight of O contained in 5g of KClO3, 1.95g. The weight we actually,obtained was 1.89g. This leaves an error of 0.06g, or a little over 4 per cent of error (0.06 / 1.95 = 0.03 +). The percentage of error, in performing this experiment, should fall within 10.
Some of the liabilities to error are as follows:—
1. Impure MnO2, which sometimes contains C. CO2 is soluble m H2O.
2. Solubility of O in water.
3. Escape of gas by leakage.
4. Moisture taken up by the gas.