Sodium hydrogen carbonate decomposes above 110°c to form sodium carbonate, water, and carbon dioxide. 2nahco3(s) na2co3(s) + h2o
1 answer:
The given chemical equation is,
When sodium carbonate is added the system tries to minimize the effect of the additional product added, by shifting the equilibrium such that the extra product is converted back to the reactants. This takes place until the equilibrium is re-established. Therefore, the equilibrium shifts towards the reactants side. The concentrations of sodium carbonate, carbon dioxide and water vapor decrease while the concentration of sodium hydrogen carbonate increases.
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Answer:
(B) 42.1%
Explanation:
The formula for the calculation of moles is shown below:
Given: For
Given mass = 15.5 g
Molar mass of = 78.0452 g/mol
<u>Moles of = 15.5 g / 78.0452 g/mol = 0.1986 moles</u>
Given: For
Given mass = 12.1 g
Molar mass of = 159.609 g/mol
<u>Moles of = 12.1 g / 159.609 g/mol = 0.0758 moles</u>
According to the given reaction:
1 mole of react with 1 mole of
0.1986 mole of react with 0.1986 mole of
Available moles of = 0.0758 moles
Limiting reagent is the one which is present in small amount. Thus, is limiting reagent. (0.0758 < 0.1986)
The formation of the product is governed by the limiting reagent. So,
1 mole of gives 1 mole of
0.0758 mole of gives 0.0758 mole of
Molar mass of = 95.611 g/mol
Mass of = Moles × Molar mass = 0.0758 × 95.611 g = 7.2473 g
Theoretical yield = 7.2473 g
Given experimental yield = 3.05 g
% yield = (Experimental yield / Theoretical yield) × 100 = (3.05/7.2473) × 100 = 42.1 %
<u>Option B is correct.</u>
Answer:
The answer to the question is;
The equilibrium constant for the reaction is 0.278
Reversibility.
Explanation:
Initial concentration = 0.500 M N₂ and 0.800 M H₂
N₂ (g) + 3·H₂ (g) ⇔ 2·NH₃ (g)
One mole of nitrogen combines with three moles of hydrogen form 2 moles of ammonia
That is 1 mole of ammonia requires 3/2 moles of H₂ and 1/2 moles of N₂
0.150 M of ammonia requires 3/2×0.150 moles of H₂ and 1/2×0.150 moles of N₂
That is 0.150 M of ammonia requires 0.225 moles of H₂ and 0.075 moles of N₂
Therefore at equilibrium we have
Number of moles of Nitrogen = 0.500 M - 0.075 M = 0.425 M
Number of moles of Hydrogen = 0.800 M - 0.225 M = 0.575 M
Number of moles of Ammonia = 0.150 M
K =
= 0.278
The kind of reaction is a reversible one as the equilibrium constant is greater than 0.01 which as general guide, all components in a reaction with an equilibrium constant between the ranges of 0.01 and 100 will be present when equilibrium is reached and the chemical reaction will be reversible.