हिंदी

1250 cc of oxygen was burnt with 300cc of ethane [C2H6]. Calculate the volume of unused oxygen formed: 2CA2HA6+7OA2⟶4COA2+6HA2O - Chemistry

Advertisements
Advertisements

प्रश्न

1250 cc of oxygen was burnt with 300cc of ethane [C2H6]. Calculate the volume of unused oxygen formed:

\[\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}\]

संख्यात्मक
Advertisements

उत्तर

\[\ce{2C2H6 + 7O2  -> 4CO2 + 6H2O}\]
2 vol         7 vol
300 cc

2 vol. of ethane requires 7 volumes of oxygen

2 vol. of ethane requires `7/2` volumes of oxygen

300 cc of ethane requires `7/2 xx 300` = 1050 cc

Total volume of oxygen = 1250 cc

Volume of oxygen used = 1050 cc

Unused oxygen = (1250 − 1050) cc

= 200 cc.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?

संबंधित प्रश्न

How does Avogadro's law explain Gay - lussac's law of combining volumes?


The reaction 4N2O + CH4 → CO2 + 2H2O + 4N2 takes place in the gaseous state. If the volumes of all the gases are measured at the same temperature, and pressure, calculate the volume of dinitrogen oxide (N2O), required to give 150cm3 of steam.


What volume of oxygen would be required to burn completely 400 ml of acetylene [C2H2]? Also calculate the volume of carbon dioxide formed.

\[\ce{2C2H2 + 5H2O -> 4CO2 + 2H2O(l)}\]


1250 cc of oxygen was burnt with 300cc of ethane [C2H6]. Calculate the volume of carbon dioxide formed:

\[\ce{2C2H6 + 7O2  -> 4CO2 + 6H2O}\]


If 6 liters of hydrogen and 4 liters of chlorine are mixed and exploded and if water is added to the gases formed, find the volume of the residual gas.


A mixture of hydrogen and chlorine occupying 36 cm3 was exploded. On shaking it with water, 4 cmof hydrogen was left behind. Find the composition of the mixture.


What volume of air (containing 20% O2 by volume) will be required to burn completely 10 cm3 of methane and acetylene?

\[\ce{CH4 + 2O2 → CO2 + 2H2O}\]

\[\ce{2C2H2 + 5O2 -> 4CO2 + 2H2O}\]


LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to atmosphere.

\[\ce{C3H8 + 5O2 → 3CO2 + 4H2O}\]

\[\ce{2C4H10 + 13O2 → 8CO2 + 10H2O}\]


The reaction: \[\ce{4N2O + CH4 -> CO2 + 2H2O + 4N2}\] takes place in the gaseous state. If all volumes are measured at the same temperature and pressure, calculate the volume of dinitrogen oxide (N2O) required to give 150 cm3 of steam.


The volumes of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temperature and pressure.

  1. Which sample of gas contains the maximum number of molecules?
  2. If the temperature and pressure of gas A are kept constant, then what will happen to the volume of A when the number of molecules is doubled?
  3. If this ratio of gas volume refers to the reactants and products of a reaction, which gas law is being observed?
  4. If the volume of A is actually 5.6 dm3 at STP, calculate the number of molecules in the actual Volume of D at STP (Avogadro's number is 6 × 1023).
  5. Using your answer from (iv), state the mass of D if the gas is dinitrogen oxide (N2O).

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×