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ratelena [41]
3 years ago
11

Aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid wat

er H2O. Suppose 0.365 g of hydrochloric acid is mixed with 0.18 g of sodium hydroxide. Calculate the maximum mass of sodium chloride that could be produced by the chemical reaction. Round your answer to 2 significant digits.
Chemistry
1 answer:
Mashcka [7]3 years ago
8 0

Answer:

0.26g of NaCl is the maximum mass that could be produced

Explanation:

Based on the reaction:

HCl + NaOH → NaCl + H₂O

<em>Where 1 mol of HCl reacts per mol of NaOH to produce 1 mol of NaCl</em>

<em />

To solve this question we need to find <em>limiting reactant. </em>The moles of limiting reactant = Moles of NaCl produced:

<em>Moles HCl -Molar mass: 36.46g/mol-:</em>

0.365g HCl * (1mol / 36.46g) = 0.010 moles HCl

<em>Moles NaOH -Molar mass: 40g/mol-:</em>

0.18g NaOH * (1mol / 40g) = 0.0045 moles NaOH

As the reaction is 1:1 and moles NaOH < moles HCl, limiting reactant is NaOH and maximum moles produced of NaCl are 0.0045 moles.

The mass of NaCl is:

<em>Mass NaCl -Molar mass: 58.44g/mol-:</em>

0.0045 moles * (58.44g/mol) =

<h3>0.26g of NaCl is the maximum mass that could be produced</h3>
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What is the balanced net precipitation reaction for BaCl2 and KOH?
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Chromium is dissolved in sulfuric acid according to the following equation: Cr + H2SO4 ⇒ Cr2 (SO4) 3 + H2
Usimov [2.4K]

Answer:

\large \boxed{\text{a)188.4 g; b) 98.67 $\, \%$}}

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:                      98.08           392.18

             2Cr + 3H₂SO₄ ⟶ Cr₂(SO₄)₃ + 3H₂

To solve the stoichiometry problem, you must

  • Use the molar mass of H₂SO₄ to convert  the mass of H₂SO₄ to moles of H₂SO₄
  • Use the molar ratio to convert moles of H₂SO₄ to moles of Cr₂(SO₄)₃
  • Use the molar mass of Cr₂(SO₄)₃ to convert moles of Cr₂(SO₄)₃ to mass of Cr₂(SO₄)₃

a) Mass of Cr₂(SO₄)₃

(i) Mass of pure H₂SO₄

\text{Mass of pure} = \text{165 g impure} \times \dfrac{\text{85.67 g pure} }{\text{100 g impure}} = \text{141.36 g pure}

(ii) Moles of H₂SO₄

\text{Moles of H$_{2}$SO}_{4} = \text{141.36 g H$_{2}$SO}_{4} \times \dfrac{\text{1 mol H$_{2}$SO}_{4}}{\text{98.08 g H$_{2}$SO}_{4}} = \text{1.441 mol H$_{2}$SO}_{4}

(iii) Moles of Cr₂(SO₄)₃

The molar ratio is 1 mol Cr₂(SO₄)₃:3 mol H₂SO₄ \text{Moles of Cr$_{2}$(SO$_{4}$)}_{3} = \text{1.441 mol H$_{2}$SO}_{4} \times \dfrac{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}}{\text{3 mol H$_{2}$SO}_{4}} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3}

(iv) Mass of Cr₂(SO₄)₃ \text{Mass of Cr$_{2}$(SO$_{4}$)}_{3} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3} \times \dfrac{\text{392.18 g Cr$_{2}$(SO$_{4}$)}_{3}}{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}} = \textbf{188.4 g Cr$_{2}$(SO$_{4}$)}_{3}\\\text{The mass of Cr$_{2}$(SO$_{4}$)$_{3}$ formed is $\large \boxed{\textbf{188.4 g}}$}

b) Percentage yield

It is impossible to get a yield of 485.9 g. I will assume you meant 185.9 g.

\text{Percentage yield} = \dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \, \% = \dfrac{\text{185.9 g}}{\text{188.4 g}} \times 100 \, \% = \mathbf{98.67 \, \%}\\\\\text{The percentage yield is $\large \boxed{\mathbf{98.67 \, \%}}$}

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4 years ago
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