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N(2) and H(2) react with to produce ammo...

`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?

Text Solution

Verified by Experts

Reaction: `underset("1 mol=28 g")(N_(2)(g)) + underset(" 3 mol of 6g")(3H_(2)(g)) to underset("2 mol=34g")(2NH_(3)(g))`
(1) 28g of nitrogen reacts with 6g of hydrogen.
`therefore`1g of nitrogen reacts with (6/28)g of hydrogen.
`therefore` 2000g of nitrogen reacts with `(6)/(28)xx2000=428.57g` of hydrogen. Therefore, nitrogen is the limiting reagent and excess amount of hydrogen is present.
28g of nitrogen produces 34g of `NH_(3)`
`therefore`2000g of `N_(2)` produces `(34)/(28)xx2000=2428.57g` of `NH_(3)`
(2) Yes, Excess hydrogen does not take part in the reaction and remains unchanged.
(3) Amount of hydrogen that does not take part in the reaction`=(1000-428.57)=571.43g`
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Knowledge Check

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