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Oksi-84 [34.3K]
3 years ago
15

Lithium and nitrogen react in a combination reaction to produce lithium nitride:6Li (s) + N2 (g) â 2Li3N (s)In a particular expe

riment, 4.50-g samples of each reagent are reacted. The theoretical yield of lithium nitride is ________ g.Lithium and nitrogen react in a combination reaction to produce lithium nitride:6Li (s) + N2 (g) 2Li3N (s)In a particular experiment, 4.50-g samples of each reagent are reacted. The theoretical yield of lithium nitride is ________ g.A) 3.76B) 22.6C) 4.53D) 7.52E) 11.2
Chemistry
1 answer:
Mademuasel [1]3 years ago
3 0

Answer:

The answer to your question is letter D

Explanation:

Data

mass of Li = 4.5 g

mass of N₂ = 4.5 g

limiting reactant = ?

Reaction

                         6Li (s)   +   N₂ (g)   ⇒  2Li₃N

This reaction is balanced

Process

1.- Calculate the molecular mass of the reactants

Li = 7 x 6 = 42g

N = 14 x 2 = 28 g

Calculate the limiting reactant using proportions

Theoretical proportion    Li / N = 42 / 28 = 1.5

Experimental proportion   Li/N  = 4.5/4.5 = 1

As the experimental proportion was lower than the theoretical, the limiting reactant is Lithium

2.- Calculate the theoretical yield of Li₃N

                        42 g of lithium --------------  70 g of Li₃N

                        4.5 g of lithium -------------   x

                        x = (4.5 x 70) / 42

                        x = 315 / 42

                     <u>   x = 7.5 %</u>

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2 years ago
Hc2h3o2(aq)+h2o(l)⇌h3o+(aq)+c2h3o−2(aq) kc=1.8×10−5 at 25∘c part a if a solution initially contains 0.260 m hc2h3o2, what is the
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Answer : The equilibrium concentration of H_3O^+ at 25^oC is, 2.154\times 10^{-3}m.

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Let, the 'x' mol/L of H_3O^+ are formed and at same time 'x' mol/L of HC_2H_3O_2 are also formed.

The equilibrium reaction is,

                  HC_2H_3O_2(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+C_2H_3O_2^-(aq)

Initially                0.260 m                       0                 0

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The expression for equilibrium constant for a given reaction is,

K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

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By rearranging the terms, we get the value of 'x'.

x=2.154\times 10^{-3}m

Therefore, the equilibrium concentration of H_3O^+ at 25^oC is, 2.154\times 10^{-3}m.

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