70.8 g of CO is exposed 20.4 g of Hy and a reaction takes place. CO + 2H2 – CH3OH 1. How many grams of methyl hydroxide (CH3OH)

Question

70.8 g of CO is exposed 20.4 g of Hy and a reaction takes place.
CO + 2H2 – CH3OH
1. How many grams of methyl hydroxide (CH3OH) is produced?
2. What is the limiting reactant?
3. How much of the excess reactant is left unused?

literally what does any of this mean​

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Dâu 5 years 2021-08-19T12:51:29+00:00 1 Answers 22 views 0

Answers ( )

  1. minhkhang
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    2021-08-19T12:52:48+00:00
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    Answer:

    1. 80.9 g of CH₃OH

    2. CO

    3. 5.14moles

    Explanation:

    The reaction is:

    CO  +  2H₂  →  CH₃OH

    First place, we determine the moles of each reactant.

    70.8 g . 1mol/ 28g = 2.53 moles of CO

    20.4 g . 1mol / 2g = 10.2 moles of H₂

    There is too much hydrogen, so the limiting reactant might be the carbon monoxide.

    2 moles of H₂ react to 1 mol of CO

    Then, 10.2 moles of H₂ may react to (10.2 . 1) /2 = 5.1 moles

    We don’t have enough CO, so it’s ok that CO is the limiting. (We have 2.53 moles, we need 5.1 moles)

    In the other hand, hydrogen is the excess reactant.

    1 mol of CO react to 2 moles of hydrogen

    2.53 moles may react to (2.53 . 2) /1  =5.06 moles

    We have 10.2 moles, and we need 5.06 moles. Then (10.2 – 5.06) = 5.14 moles remains after the reaction goes complete.

    Now that we know the limiting reactant, we can determine the grams of produced methanol. Ratio is 1:1

    1 mol of CO produces 1 mol of CH₃OH

    2.53 moles of CO must produce 2.53 moles of CH₃OH

    We convert moles to mass: 2.53 mol . 32 g/mol = 80.9 g

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