Is this equation balanced?nh4oh+hc2h3o2=nh4c2h3o2+h2o
2 answers:
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Answer : The percent yield of the reaction is, 76.34 %
Explanation : Given,
Pressure of and
= 25.0 atm
Temperature of and
=
Volume of = 1050 L per min
Volume of = 1550 L per min
R = gas constant = 0.0821 L.atm/mole.K
Molar mass of = 30 g/mole
First we have to calculate the moles of and
by using ideal gas equation.
For :
For :
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 1 mole of react with 1 mole of
So, 611.34 mole of react with 611.34 mole of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of .
As, 1 mole of react to give 1 mole of
As, 611.34 mole of react to give 611.34 mole of
Now we have to calculate the mass of .
The theoretical yield of = 18340.2 g
The actual yield of = 14.0 kg = 14000 g (1 kg = 1000 g)
Now we have to calculate the percent yield of
Therefore, the percent yield of the reaction is, 76.34 %
Answer:
Explanation:
Given parameters;
pH = 8.74
pH = 11.38
pH = 2.81
Unknown:
concentration of hydrogen ion and hydroxyl ion for each solution = ?
Solution
The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.
It is graduated from 1 to 14
pH = -log[H₃O⁺]
pOH = -log[OH⁻]
pH + pOH = 14
Now let us solve;
pH = 8.74
since pH = -log[H₃O⁺]
8.74 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 1.82 x 10⁻⁹mol dm³
pH + pOH = 14
pOH = 14 - 8.74
pOH = 5.26
pOH = -log[OH⁻]
5.26 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] = 5.5 x 10⁻⁶mol dm³
2. pH = 11.38
since pH = -log[H₃O⁺]
11.38 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 4.17 x 10⁻¹² mol dm³
pH + pOH = 14
pOH = 14 - 11.38
pOH = 2.62
pOH = -log[OH⁻]
2.62 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] =2.4 x 10⁻³mol dm³
3. pH = 2.81
since pH = -log[H₃O⁺]
2.81 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 1.55 x 10⁻³ mol dm³
pH + pOH = 14
pOH = 14 - 2.81
pOH = 11.19
pOH = -log[OH⁻]
11.19 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] =6.46 x 10⁻¹²mol dm³