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3 years ago

How many liters of 4.0 M NaOH solution will react with 1.2 mol H2SO4? (Remember to balance the equation.)

Chemistry

2 answers:

valina [46]3 years ago

8 0

The answer to this is D

nydimaria [60]3 years ago

6 0

Answer:

The answer is D. 0.60 L

Explanation:

The balanced reaction equation including states of matter is;

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

More simple:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Now, we can see from this reaction equation that the mole ratio of NaOH to H2SO4 is 2:1

Number of moles of H2SO4 reacted = 1.2 moles

Hence;

2 moles of NaOH reacts with 1 mole of H2SO4

x moles of NaOH reacts with 1.2 moles of H2SO4

x = 2 * 1.2/1 = 2.4 moles of NaOH

Recall that;

Number of moles = Concentration * Volume

Volume = number of moles/concentration

Volume of NaOH is obtained from;

Volume = 2.4 moles/ 4.0 M

Volume = 0.60 L

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Explanation:

Volume of cylindrical reservoir = V

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Mass of water cylindrical reservoir = m

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1.6 kilogram of fluorine per million kilograms of water. (Given)

Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1.6 kg of fluorine

Then $15,896,250 kg$ of water have x mass of fluorine:

$\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}$

$x=\frac{1.6 kg}{1000,000}\times 15,896,250 kg=25.434 kg$

15,896,250 kg water of contains mass 25.434 kg of fluorine.

25.434 kg = 25434 g

25,434 grams of fluorine should be added to give 1.6 ppm.

Percentage of fluorine in hydrogen hexafluorosilicate :

Molar mass hydrogen hexafluorosilicate = 144 g/mol

$F\%=\frac{6\times 19 g/mol}{144 g/mol}\times 100=79.16\%$

Total mass of hydrogen hexafluorosilicate = m'

$79.16\%=\frac{25,434 g}{m'}\times 100$

m' = 32,127.02 g

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