The moles of `O_(2)` required for reacting with `6*8` g ammonia
`(...NH_(3)+....O_(2)to....NO+....H_(2)O)` is

The moles of O_(2) required for reacting with 6.8 gm of ammonia. (4NH_(3)+5O_(2) rarr 4NO+6H_(2)O) is

Statement-1 : When 1 mole of NH_3 and 1 mole of O__(2) are made to react, all the NH_(3) may be consumed, if reactions is : NH_(3)(g)+O_(2)(g)toNO(g)+H_(2)O(l) Statement-2 : Oxygen is limiting reagent.

According to the equation N_(2)O_(3)(g) + 6H_(2)(g) rArr 2NH_(3)(g) + 3H_(2)O(g) how many moles of NH_(3)(g) could be formed from the reaction of 0.22 mol of N_(2)O_(3) (g) with 0.87 mol of H_(2)(g) ?

How many coulombs are required for the oxidation of 1 mole of H_(2)O_(2) to O_(2) ?

The number of moles of H_(2) which is required to produce 10 moles of NH_(3) in the following reaction is aH_(2)(g)+bNO_(2)(g)tocNH_(3)(g)+dH_(2)O(g)

In the reaction 4NH_(3)+3O_(2(g)) to 2N_(2(g))+6H_(2)O_((g))

The number of moles of Fe_(2)O_(3) formed when 5.6 lit of O_(2) reacts with 5.6g of Fe?