Item 3 Part A Carbon monoxide gas reacts with hydrogen gas to form methanol via

Question

Item 3 Part A Carbon monoxide gas reacts with hydrogen gas to form methanol via the following reaction: Identify the limiting reactant and determine the theoretical yield of methanol in grams. Express your answer with the appropriate units. CO(g) + 2H2 (g)-CH,011(g) A 1.65 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 393 mmHg Value Units You may want to reference (pages 444-446) Section 10.10 while completing this problem. Submit My Answers Give Up Provide Feedback Continue

Explanation / Answer

PCO = 232mmHg   = 232/760   = 0.305atm

V   = 1.65L

T = 305K

PV = nRT

n   = PV/RT

   = 0.305*1.65/0.0821*305   = 0.02 moles of CO

PH2 = 393mmHg   = 393/760   = 0.517atm

V   = 1.65L

T = 305K

PV = nRT

n   = PV/RT

    =0.517*1.65/0.0821*305     = 0.034 moles of H2

CO + 2H2 --------------------> CH3OH

1 mole of CO react with 2 moles of H2

0.02 moles of Co react with = 2*0.02/1 = 0.04 moles of H2

H2 is limiting reactant

CO + 2H2 --------------------> CH3OH

2 moles of H2 react with CO to gives 1 moles of CH3OH

0.034 moles of H2 react with CO to gives = 0.034*1/2    = 0.017 moles of CH3OH

mass of CH3OH   = no of moles * gram molar mass

                             = 0.017*32   = 0.544g

theoritical yield of CH3OH = 0.544g >>>.answer

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