[By Lussac’s law]
2C2H2 | + | 5O2 | → | 4CO2 | + | 2H2O |
2 vol | : | 5 vol | 4 vol | : | 2 vol |
To calculate the volume of ethyne gas
CO2 | : | C2H2 |
4 | : | 2 |
8.4 | : | x |
⇒ x = 2\over 4 × 8.4 = 4.2 dm3
Hence, volume of ethyne gas required = 4.2 dm3.
[By Lussac’s law]
2C2H2 | + | 5O2 | → | 4CO2 | + | 2H2O |
2 vol | : | 5 vol | 4 vol | : | 2 vol |
To calculate the volume of ethyne gas
CO2 | : | C2H2 |
4 | : | 2 |
8.4 | : | x |
⇒ x = 2\over 4 × 8.4 = 4.2 dm3
Hence, volume of ethyne gas required = 4.2 dm3.
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