What is the maximum mass of `H_(2)SO_(4)` that can be obtained from a 100 gm oleum sample labelled as 110%

What is the maximum mass of CO_(2) that can be obtained by combustion of 4.4 gm propane-

50 gm of 109 % oelum is mixed with 50 gm of another 118 % oleum. Calculate the maximum weight of H_(2)SO_(4) which can be obtained from the resulting mixture. (Fill the OMR after multiplying with 10)

What mass of H_(2)SO_(4) contains 32g oxygen ? .

An oleum sample is labelled as 118%, Calculate Mass of H_(2)SO_(4) in 100 gm oleum sample

The percentage labelling (mixture of H_(2)SO_(4) and SO_(3)) refers to the total mass of pure H_(2)SO_(4) . The total amount of H_(2)SO_(4) found after adding calculated amount of water to 100 g oleum is the percentage labelling of oleum. Higher the percentage lebeling of oleum higher is the amount of free SO_(3) in the oleum sample. What is the amount of free SO_(3) in an oleum sample labelled as '118%'.

sulphuric acid is produced when Sulphur dioxide reacts with oxygen and water in the presence of a catalyst : 2SO_(2)(g) + O_(2) (g) + 2H_(2)O(l) to2H_(2)SO_(4) . If 5.6 mol of SO_(4) reacts with 4.8 mol of O_(2) and a large excess of water , what is the maximum number of moles of H_(2)SO_(4) that can be obtained ?

Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as 109% H_(2)SO_(4) means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) What is the % of free SO_(3) in an oleum that is labelled as 104.5% H_(2)SO_(4) ?

Find the % labelling of 100 gm oleum sample if it contains 20 gm SO_(3)

Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as ' 019% H_(2)SO_(4) ' means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) 1 g of oleum sample is diluted with water. The solution required 54 mL of 0.4 N NaOH for complete neutralization. The % free SO_(3) in the sample is : (a)74 (b)26 (c)20 (d)None of these

Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as ' 109% H_(2)SO_(4) ' means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) If excess water is added into a bottle sample labelled as "112% H_(2)SO_(4) " and is reacted with 5.3 g NaCO_(3) then find the volume of CO_(2) evolved at 1 atm pressure and 300 K temperature after the completion of the reaction :