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Andrei [34K]
3 years ago
7

Using this balanced equation: 2 NaOH + H2SO4 —> H2O + Na2SO4

Chemistry
1 answer:
mars1129 [50]3 years ago
7 0
<h3>Answer:</h3>

266.325 g

<h3>Explanation:</h3>

We are given the balanced equation;

2NaOH + H₂SO₄ → H₂O + Na₂SO₄

  • Mass of NaOH as 150 g

We are required to determine the mass of Na₂SO₄ that will be formed.

<h3>Step 1: Determine the number of moles of NaOH</h3>

Moles = Mass ÷ molar mass

Molar mass of NaOH is 40.0 g/mol

Therefore;

Moles of NaOH = 150 g ÷ 40 g/mol

                          = 3.75 moles

<h3>Step 2: Determine the number of moles of sodium sulfate formed</h3>
  • From the equation 2 moles of NaOH reacts with sulfuric acid to form 1 mole of sodium sulfate.
  • Therefore; mole ratio of NaOH : Na₂SO₄ is 2 : 1

Thus, moles of Na₂SO₄ = Moles of NaOH ÷ 2

                                      = 3.75 moles ÷ 2

                                     = 1.875 moles

<h3>Step 3: Determine the mass of Na₂SO₄ produced.</h3>

we know that;

Mass = Moles × Molar mass

Molar mass of Na₂SO₄ is 142.04 g/mol

Therefore;

Mass of Na₂SO₄ = 1.875 moles × 142.04 g/mol

                           = 266.325 g

Thus, the mass of sodium sulfate formed 266.325 g

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Answer:

0.8749 grams of hydrogen gas was formed from the reaction.

Explanation:

P = Pressure of hydrogen gad= 744 Torr = 0.98 atm

(1 atm = 760 Torr)

V = Volume of hydrogen gas= 11 L

n = number of moles of hydrogen gas= ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 27.0 °C = 300.15 K

Putting values in above equation, we get:

Using an ideal gas equation:

PV=nRT

n=\frac{PV}{RT}=\frac{0.98 atm\times 11 L}{0.0821 L atm/mol K\times 300.15 K}

n = 0.4374 moles

Mass of 0.4374 moles of hydrogen gas:

0.4374 mol × 2 g/mol = 0.8749 g

0.8749 grams of hydrogen gas was formed from the reaction.

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3 years ago
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2 years ago
Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili
Katen [24]

<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}

\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M

\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

                         H_2+I_2\rightleftharpoons 2HI

<u>Initial:</u>                0.1    0.1

<u>At eqllm:</u>          0.1-x   0.1-x   2x

Evaluating the value of 'x'

\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M

The expression of K_c for above equation follows:

K_c=\frac{[HI]^2}{[H_2][I_2]}

[HI]_{eq}=2x=(2\times 0.079)=0.158M

[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M

[I_2]_{eq}=0.0210M

Putting values in above expression, we get:

K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61

Hence, the value of equilibrium constant for the given reaction is 56.61

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