280 g of a mixture containing `CH_(4) and C_(2)H_(6)` in `5:2` molar ratio is burnt in presence of excess of oxygen. Calculate total moles of `CO_(2)` produced.

A mixture containing 3 moles each of C_(4)H_(8) and C_(6)H_(6) undergoes combustion with O_(2) to form CO_(2) and H_(2)O . Calculate total mass of CO_(2) produced

A mixture containing 3 moles each of C_(4)H_(8) and C_(6)H_(6) undergoes combustion with O_(2) to form CO_(2) and H_(2)O . Calculate total mass of CO_(2) produced

A 10 gram sample of natural gas containing CH_(4) and C_(2)H_(4) was burnt in excess of oxygen to give 29.0 "grams" of CO_(2) and some water. How many games of water are formed :

A 10 gram sample of natural gas containing CH_(4) and C_(2)H_(4) was burnt in excess of oxygen to give 29.0 "grams" of CO_(2) and some water. How many games of water are formed :

A 10 gram sample of natural gas containing CH_(4) and C_(2)H_(4) was burnt in excess of oxygen to give 29.0 "grams" of CO_(2) and some water. How many grams of water are formed :

A gaseous mixture containing He, CH_(4) and SO_(2) in 1:2:3 mole ratio, calculate the molar ratio of gases effusing out initially.

A gaseous mixture containing He, CH_(4) and SO_(2) in 1:2:3 mole ratio, calculate the molar ratio of gases effusing out initially.

A gaseous mixture containing He, CH_(4) and SO_(2) in 1:2:3 mole ratio, calculate the molar ratio of gases effusing out initially.

A mixture containing 3 moles each of C_4H_8 and C_6H_6 undergoes complete combustion with O_2 to form CO_2 and H_2O Calculate total mass of CO_2 produced :

The percentage by volume of C_(3)H_(8) in a gaseous mixture of C_(3)H_(8),CH_(4) and CO is 20. When 100mL of the mixture is burnt in excess of O_(2) the volume of CO_(2) produced is :