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Elena L [17]
3 years ago
10

You are given solutions of HCl and NaOH and must determine their concentrations. You use 30.0 mL of NaOH to neutralize 100.0 mL

of HCl, and 20.0 mL of NaOH to neutralize 50.0 mL of 0.361 M H2SO4. What is the concentration of the HCl solution? Enter your answer in units of M to three significant figures.
Chemistry
1 answer:
valina [46]3 years ago
7 0

Answer:

0.542M HCl

Explanation:

The reaction of H₂SO₄ with NaOH is:

H₂SO₄ + 2 NaOH → 2H₂O + Na₂SO₄

<em>Where 1 mole of acid reacts with 2 moles of NaOH</em>

Moles of H₂SO₄ are:

0.0500L × (0.361mol / L) = 0.01805 moles H₂SO₄

Thus, moles of NaOH that neutralize this acid are:

0.01805 moles H₂SO₄ × (2 mol NaOH / 1 mol H₂SO₄) = 0.0361 moles NaOH

And concentration is:

0.0361 moles NaOH / 0.0200L = <em>1.805M</em>

And, reaction of NaOH with HCl is:

NaOH + HCl → H₂O + NaCl

<em>Where 1 mole of NaOH reacts per mole of NaOH</em>

As you use 30.0mL = 0.0300L of NaOH to neutralize the HCl acid, moles of acid are:

0.0300L × (1.805mol / L) = 0.05415 moles NaOH = moles HCl

In 0.1000L:

0.05415 moles HCl / 0.1000L = <em>0.542M HCl</em>

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Answer:

0.4694 moles of CrCl₃

Explanation:

The balanced equation is:

Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)

The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.

The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:

MCr = 52 g/mol

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MO = 16 g/mol

So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.

The number of moles is the mass divided by the molar mass, so:

n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.

For the stoichiometry:

1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃

0.2347 mol of Cr₂O₃----------- x

By a simple direct three rule:

x = 0.4694 moles of CrCl₃

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Explanation:

Firstly we analyse data.

12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.

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We need to determine, the volume of solution for the concentration

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Volume =  100 g / 1.05 g/mL → 95.24 mL

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Let's finish this by a rule of three.

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