Question

If 48 g of Mg metal is reacted with 34g$N{H_3}$ gas. What is the mass of $M{g_3}{N_2}$ produced?
A. 200/3 g
B. 100/3 g
C. 400/3 g
D. 150/3 g

Answer 1

Hint: A balanced equation for the reaction carried above is:
     $3Mg + 2N{H_3} \to M{g_3}{N_2} + 3{H_2}$

Complete step by step solution:
From Avogadro’s law it is known that equal volumes of all gases contain equal numbers of molecules under similar conditions. Now since one mole of molecules of all gases contain the same number (6.022$ \times {10^{23}}$ ) of molecules, therefore they occupy the same volume under similar pressure and temperature. The mass of one mole of atoms is exactly equal to the atomic mass in grams of that element.
In the equation above we see that 3 moles of magnesium react with 2 moles of ammonia gas to produce 3 moles of hydrogen gas and$M{g_3}{N_2}$. We know that atomic mass of magnesium is 24g which contribute to one mole of magnesium and atomic mass of ammonia gas is (atomic mass of nitrogen + 3*atomic mass of hydrogen) which is 17g. Therefore 1 mole of ammonia gas.
According to the question if 48 g of Mg is reacting therefore 2 moles of Mg reacts with 2 moles of ammonia but in the reaction 3 moles of Mg produces 1 mole of magnesium nitride that is 100g of magnesium nitride is formed
(3 moles of Mg) 72g of Mg produces= 100g of magnesium nitride (1 mole of magnesium nitride)
     48g of Mg will produce=$\dfrac{{100 \times 48}}{{72}} = \dfrac{{200}}{3}$ g
Hence the correct option is option A.

Additional information: A limiting reagent in a reaction is the species supplied in an amount smaller than that required by the stoichiometric relation between the reactants. In the above reaction limiting reagent is magnesium.

Note: Balance the equation before solving such questions on stoichiometry.