Carbon monoxide gas reacts with hydrogen gas to form methanolvia the following reaction:CO(g)2H2 (g)->CH3 OH(g)Part AA 1.50 L reaction vessel, initially at 305 K, contains carbonmonoxide gas at a partial pressure of 232 mmHg and hydrogengas at a partial pressure of 367 mmHgDetermine the theoretical yield of methanol in grams.Express your answer with the appropriate units.?You may want to reference (Pages 444 - 446) Section 10.10 whilecompleting this problemValueUnitsRequest AnswerSubmit

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Asked Nov 14, 2019
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Carbon monoxide gas reacts with hydrogen gas to form methanol
via the following reaction:
CO(g)2H2 (g)->CH3 OH(g)
Part A
A 1.50 L reaction vessel, initially at 305 K, contains carbon
monoxide gas at a partial pressure of 232 mmHg and hydrogen
gas at a partial pressure of 367 mmHg
Determine the theoretical yield of methanol in grams.
Express your answer with the appropriate units.
?
You may want to reference (Pages 444 - 446) Section 10.10 while
completing this problem
Value
Units
Request Answer
Submit
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Carbon monoxide gas reacts with hydrogen gas to form methanol via the following reaction: CO(g)2H2 (g)->CH3 OH(g) Part A A 1.50 L reaction vessel, initially at 305 K, contains carbon monoxide gas at a partial pressure of 232 mmHg and hydrogen gas at a partial pressure of 367 mmHg Determine the theoretical yield of methanol in grams. Express your answer with the appropriate units. ? You may want to reference (Pages 444 - 446) Section 10.10 while completing this problem Value Units Request Answer Submit

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Step 1

The number of moles (n1) of CO (g) inside the 1.50 L reaction vessel at 305 K is calculated as shown in equation (1) where R is the gas constant and P1 is the partial pressure of CO. The values for R and P1 are substituted in equation (1). The number of moles (n1) of CO (g) inside the 1.50 L reaction vessel at 305 K is 0.018 mol.

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...1) Рx1.50L %3D nRx305 K 232 mmHg x150 L п 62.364 mmHg Kmol" x 305 K =0.018 mol

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Step 2

The number of moles (n2) of H2 (g) inside the 1.50 L reaction vessel at 305 K is calculated as shown in equation (2) where R is the gas constant and P2 is the partial pressure of H2. The values for R and P1 are substituted in equation (2). The number of moles (n2) of H2 (g) inside the 1.50 L reaction vessel at 305 K is 0.029 mol.

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(2) Р. х1.50 L %3D n,Rx305 K 367 mmHg x150 È п, 62.364 LmmHg mol1x305 K 0.029 mol

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Step 3

According to the given chemical reaction, 0.018 mol of CO (g) require 0.036 mol of H2 (g) thus, H2 (g) is the limiting reagent and the amount of formation of CH3OH depends upon th...

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