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vlada-n [284]
2 years ago
6

A reaction vessel contains nh3, n2, and h2 at equilibrium at a certain temperature. the equilibrium concentrations are [n2] = 0.

31 m, [h2] = 1.51 m, and [nh3] = 0.75 m. calculate the equilibrium constant, kc, if the reaction is represented as
Chemistry
1 answer:
Nat2105 [25]2 years ago
5 0
The balanced equation for the reaction is,
          N₂(g) + 3H₂(g) ⇄ 2NH₃(g)

Since the given concentrations are taken at the equilibrium state, we can use those directly for the calculation.

Kc = [NH₃(g)]² / ([N₂(g)] x [H₂(g)]³)
Kc = (0.75 M)² / ((0.31 M) x (<span>1.51 M</span>)³)
Kc = 0.527 M⁻²
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B) hyperbolic curve; saturated with substrate

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2 years ago
We have an object with a density of 620 g/ cm3 and a volume of 75 cm3. What is the mass of this object?
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If 10.0 grams of NaHCO3 is added to 10.0 g of HCl, determine the efficiency of baking soda as an antacid if 6.73 g of NaCl was p
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percentage yield of NaCl = 96.64%

Explanation:

The reaction was between NaHCO3 and HCl .The chemical equation can be represented below:

NaHCO3 + HCl → NaCl + H2O + CO2 . The balance equation is

NaHCO3 + HCl → NaCl + H2O + CO2

The question ask us to calculate the percentage yield of NaCl.

The efficiency of NaHCO3 as an antacid , the limiting reactant is NaHCO3

as

1 mole of NaHCO3 produces 1 mole of NaCl

Therefore,

molar mass of NaHCO3 = 23 +1 + 12 + 48 = 84 g

molar mass of NaCl = 23 + 35.5 = 58.5 g

1 mole of NaHCO3 = 84 g

1 mole of NaCl  = 58.5 g

since 84 g of NaHCO3 produces 58.5 g of NaCl

10 g of NaHCO3 will produce ? grams of NaCl

cross multiply

Theoretical yield of NaCl = (10 × 58.5)/84

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percentage yield of NaCl = actual yield/theoretical yield × 100

percentage yield of NaCl = 6.73/6.9642857143 × 100

percentage yield of NaCl = 673/6.9642857143

percentage yield of NaCl = 96.635897436%

percentage yield of NaCl = 96.64%

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