**Chemical Reaction Analysis: Carbon Monoxide and Hydrogen Gas to Methanol****Reaction Overview:**Carbon monoxide gas reacts with hydrogen gas to form methanol:\[ \text{CO (g) + 2H}_2 \text{(g) } \longrightarrow \text{ CH}_3\text{OH (g)} \]**Experimental Setup:**- **Volume of Reaction Vessel:** 3.0 L- **Temperature:** 325 K- **Partial Pressure of CO:** 201 mmHg- **Partial Pressure of H₂:** 387 mmHg**Problem Statement:**Identify the limiting reactant and determine the theoretical yield of methanol in grams. Show all your work.(Note: Ensure to use the ideal gas law for calculations and molar masses for conversion to grams.)

The considered reaction is, CO (g) + 2H2 (g)→CH3OH (g)Volume of vessel (V)=3.0 LTemperature (T)=325 KTotal pressure (P)=201+387=588 mmHg=0.7737 atmGas constant (R)=0.08205 L.atm.mol-1.K-1Total no. of moles of gases =PVRT=0.7737×30.08205×325=0.087Partial pressure of gas=Mole ratio×Total pressurePartial pressure of CO=201 mmHg=0.2645 atm0.2645=Moles of CO0.087×0.7737No. of moles of CO taken=0.0297Partial pressure of H2=387 mmHg=0.5092 atm0.5092=Moles of H20.087×0.7737Moles of H2 taken=0.0573In this reaction, one mole of CO can react with 2 moles of H2. Hence, 0.0297 moles of CO can react with 0.0594 moles of H2. However, only 0.0573 moles of H2 are available. Hence, some amounts of unreacted CO will be present apart from the methanol formed in the reaction. Hence, H2 gas is the limiting reactant. If the reaction goes to completion and all the available H2 is reacted with half its amount of CO, then, using the molar ratio of products and reactants obtained from the balanced chemical equation, the theoretical yield can be predicted. Maximum no. of moles of methanol that can be formed=No. of moles of H22=0.05732=0.02865Theoretical yield=No. of moles ×Molar mass of methanol=0.02865×32.04=0.91795 g≃0.918 g

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Carbon monoxide gas reacts with hydrogen gas to form methanol: CO (g) + 2H2 (g) --> CH3OH (g) A 3.0 L reaction vessel, at 325 K, contains carbon monoxide gas at a partial pressure of 201 mmHg and hydrogen gas at a partial pressure of 387 mmHg. Identify the limiting reactant and determine the theoretical yield of methanol in grams. Show all your work

Carbon monoxide gas reacts with hydrogen gas to form methanol: CO (g) + 2H2 (g) --> CH3OH (g) A 3.0 L reaction vessel, at 325 K, contains carbon monoxide gas at a partial pressure of 201 mmHg and hydrogen gas at a partial pressure of 387 mmHg. Identify the limiting reactant and determine the theoretical yield of methanol in grams. Show all your work

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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Mole Concept

Mole concept is a method to determine the amount of any substance that consists of some elemental particles using a grouping unit called ‘mole’. The mole concept enables one to handle the large amount of small elementary particles. The mole concept can be used in many calculations related to mole calculation, molar amount, mole ratio, and so on, which are significant in chemistry.

Question

**Chemical Reaction Analysis: Carbon Monoxide and Hydrogen Gas to Methanol**

**Reaction Overview:**
Carbon monoxide gas reacts with hydrogen gas to form methanol:
\[ \text{CO (g) + 2H}_2 \text{(g) } \longrightarrow \text{ CH}_3\text{OH (g)} \]

**Experimental Setup:**
- **Volume of Reaction Vessel:** 3.0 L
- **Temperature:** 325 K
- **Partial Pressure of CO:** 201 mmHg
- **Partial Pressure of H₂:** 387 mmHg

**Problem Statement:**
Identify the limiting reactant and determine the theoretical yield of methanol in grams. Show all your work.

(Note: Ensure to use the ideal gas law for calculations and molar masses for conversion to grams.)

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