**Problem Description:**A mixture of 20.0 g of phosphorus (P) and 78.8 g of chlorine (Cl₂) reacts completely to form phosphorus trichloride (PCl₃) and phosphorus pentachloride (PCl₅) as the only products.**Part A:**Find the mass of PCl₃ that forms.**Instructions:**Express your answer with the appropriate units.**Input Box:**- Answer submitted: 38.2 g**Buttons:**- [Submit]- Previous Answers- Request Answer**Feedback:**- Message: Incorrect; Try Again; One attempt remaining**Explanation:**The interactive module requires calculating the mass of PCl₃ produced in a chemical reaction where phosphorus and chlorine are the reactants. The user inputted an answer of 38.2 grams, but it was marked incorrect. Only one attempt remains to input the correct mass.

To calculate the mass of PCl3 formed when 20.0 g of P reacts completely with 78.8 g of Cl2, we can…

Answered: A mixture of 20.0 g of P and 78.8 g of…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

**Problem Description:**

A mixture of 20.0 g of phosphorus (P) and 78.8 g of chlorine (Cl₂) reacts completely to form phosphorus trichloride (PCl₃) and phosphorus pentachloride (PCl₅) as the only products.

**Part A:**

Find the mass of PCl₃ that forms.

**Instructions:**

Express your answer with the appropriate units.

**Input Box:**

- Answer submitted: 38.2 g

**Buttons:**

- [Submit]
- Previous Answers
- Request Answer

**Feedback:**

- Message: Incorrect; Try Again; One attempt remaining

**Explanation:**

The interactive module requires calculating the mass of PCl₃ produced in a chemical reaction where phosphorus and chlorine are the reactants. The user inputted an answer of 38.2 grams, but it was marked incorrect. Only one attempt remains to input the correct mass.

Expert Solution

Step 1: Provided informatio

To calculate the mass of PCl₃ formed when 20.0 g of P reacts completely with 78.8 g of Cl₂, we can apply stoichiometry. The balanced chemical equation for the reaction as

$bold italic 2 bold italic P bold italic plus bold italic 3 bold italic C bold italic l subscript bold italic 2 bold italic rightwards arrow bold italic 2 bold italic P bold italic C bold italic l subscript bold italic 3 bold italic space bold italic rightwards arrow bold italic left parenthesis bold italic a bold italic right parenthesis$ $bold italic space bold italic 2 bold italic P bold italic plus bold italic 5 bold italic C bold italic l subscript bold italic 2 bold italic rightwards arrow bold italic 2 bold italic P bold italic C bold italic l subscript bold italic 5$

we know that; $bold italic A bold italic t bold italic o bold italic m bold italic i bold italic c bold italic space bold italic m bold italic a bold italic s bold italic s bold italic space bold italic o bold italic f bold italic space bold italic P bold italic equals bold italic 30 bold italic. bold italic 974 bold italic space bold italic g bold italic. bold italic m bold italic o bold italic l to the power of bold italic minus bold italic 1 bold italic space bold italic space bold italic space bold italic space end exponent bold italic & bold italic space bold italic m bold italic o bold italic l bold italic a bold italic r bold italic space bold italic m bold italic a bold italic s bold italic s bold italic space bold italic C bold italic l subscript bold italic 2 bold italic equals bold italic space bold italic 70 bold italic. bold italic 906 bold italic space bold italic g bold italic. bold italic m bold italic o bold italic l to the power of bold italic minus bold italic 1 end exponent$

Now, calculate the number of moles for each reactant as

$M o l e s space o f space P h o s p h o r u s equals fraction numerator M a s s space left parenthesis g right parenthesis space over denominator M o l a r space m a s s space left parenthesis g. m o l to the power of negative 1 end exponent right parenthesis space end fraction rightwards double arrow space fraction numerator 20.0 space g space over denominator 30.974 space g. m o l to the power of negative 1 end exponent end fraction space almost equal to space bold italic 0 bold italic. bold italic 6457 bold italic space bold italic m bold italic o bold italic l bold italic e bold italic s$
$M o l e s space o f space C l subscript 2 equals fraction numerator M a s s space left parenthesis g right parenthesis space over denominator M o l a r space m a s s space left parenthesis g. m o l to the power of negative 1 end exponent right parenthesis space end fraction rightwards double arrow space fraction numerator 78.8 space g space over denominator 70.906 space g. m o l to the power of negative 1 end exponent end fraction space almost equal to bold 1 bold. bold 112 bold space bold italic m bold italic o bold italic l bold italic e bold italic s$

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