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3 years ago

The overall reaction taking place is shown in the equation below

Chemistry

1 answer:

algol [13]3 years ago

7 0

A. The maximum possible mass of sodium that can be produced is 47.18 Kg

B. The percentage yield of the reaction is 80.6%

<h3>Determination of the mass of NaCl and Na </h3>

We'll begin by calculating the mass of NaCl that reacted and the mass of Na obtained from the balanced equation.

2NaCl –> 2Na + Cl₂

Molar mass NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl from the balanced equation = 2 × 58.5 = 117 g = 117 / 1000 = 0.117 Kg

Molar mass of Na = 23 g/mol

Mass of Na from the balanced equation = 2 × 23 = 46 g = 46 / 1000 = 0.046 Kg

SUMMARY

From the balanced equation above,

0.117 Kg of NaCl reacted to produce 0.046 Kg of Na.

<h3>A. How to determine mass of Na produced </h3>

From the balanced equation above,

0.117 Kg of NaCl reacted to produce 0.046 Kg of Na.

Therefore,

120 Kg of NaCl will react to produce = (120 × 0.046) / 0.117 = 47.18 Kg of Na

Thus, the maximum mass of Na produced is 47.18 Kg.

<h3>B. How to determine the percentage yield </h3>

Actual yield = 38.05 Kg
Theoretical yield = 47.18 Kg

Percentage yield =?

Percentage yield = (Actual / Theoretical) × 100

Percentage yield = (38.05 / 47.18) × 100

Percentage yield = 80.6%

Learn more about stoichiometry:

brainly.com/question/14735801

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4 years ago

2AgNO3 + BaCl2 → 2AgCl + Ba(NO3)2

miv72 [106K]

<h3>Answer:</h3>

4 g AgCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN] 2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂

[Given] 5.0 g AgNO₃

<u>Step 2: Identify Conversions</u>

[Reaction - Stoich] 2AgNO₃ → 2AgCl

Molar Mass of Ag - 107.87 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 5.0 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{2 \ mol \ AgCl}{2 \ mol \ AgNO_3})(\frac{143.22 \ g \ AgCl}{1 \ mol \ AgCl})$
Multiply/Divide: $\displaystyle 4.21533 \ g \ AgCl$