Ammonia is produced in accordance with this equation .
`N_(2)(g) + 3H_(2)(g) rightarrow 2NH_(3)(g)`
In a particular experiment, 0.25mol of `NH_(3)` is formed when 0.5 mol of `N_(2)` is reacted with 0.5 mol of `H_(2)`. What is the percent yield?

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) ?

20.0 kg of N_(2)(g) and 3.0 kg of H_(2)(g) are mixed to produce NH_(3)(g) . The amount of NH_(3)(g) formed is

For the reaction N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g) , DeltaH = -93.6 KJ mol^(-1) the formation of NH_(3) is expected to increase at :

For the reaction N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g) , DeltaH = -93.6 KJ mol^(-1) the formation of NH_(3) is expected to increase at :

N_(2) and H_(2) react with to produce ammonia according to the equation: N_(2)(g)+3H_(2)(g) to 2NH_(3)(g) (1) Calculate the mass of ammonia produced if 2.00xx10^(3)g" "N_(2) reacts with 1.00xx10^(3)g off H_(2) . (2) Will any of the two reactants remain unreacted ? (3) If yes, which one and what would be its mass?

Heat change for a reaction N_(2)(g) + 3H_(2) (g) to 2NH_(3)(g) is - 92.2 kj/mol. Calculate the heat of formation of one mole of a ammonia.