The mass of Mg3N2 produced, if 48 g of Mg metal reacted with 34 g of NH3 gas is
(Mg = 24, N=14, H=1)
3Mg + 2NH2→ Mg3N2 + 3H2
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Answer:
66.67g
Explanation:
In the equation given clearly we know that 72g of Mg(3*24) and 34g of NH3 reacts to give 100g of Mg3N2.
Finding the limiting reagent :
given in question: 48g of Mg reacts with 34g of NH3
=> No. of moles of My taken in 2nd case is reduced to 2mols ( actually required 3):
3mol ----> 72g (Mg)
x--------> 48g
x=( 48*3)/72= 2mols
therefore Mg is the limiting reagent.
The amount of product formed depends only on the limiting reagent.
Therefore,
72g(Mg)-----> 100g(Mg3N2)
48g. ------> x
=> x= (100*48)/72
=> x= 66.66666g=66.67g
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