Methanol (CH3OH) is made by reacting gaseous carbon monoxide (CO) and gaseous hydrogen (H2).

A. What is the theoretical yield of methanol if you react 37.5 kg of CO(g) with 4.60 kg of H2(g)?
B. You found that 1.83 x 104 g of methanol is actually produced. What is the percent yield of methanol?

A. Theoretical yield is 36800g

B. 49.73%

Explanation:

First let us generate a balanced equation for the reaction. This is illustrated below:

CO + 2H2 —> CH3OH

From the question given, we obtained the following:

Mass of CO = 37.5 kg = 37.5 x 1000 = 37500g

Mass of H2 = 4.60kg = 4.60 x 1000 = 4600g

Let us convert these Masses to mol

For CO:

Molar Mass of CO = 12 + 16 = 28g/mol

Mass of CO = 37500g

Number of mole of CO = 37500/28 = 1339.3moles

For H2:

Molar Mass of H2 = 2x1 =2g/mol

Mass of H2 = 4600g

Number of mole of H2 = 4600/2 =

2300moles

Now we can see from the equation above that for every 2moles of H2, 1mole of CO is required. Therefore, 2300moles of H2 will require = 2300/2 = 1150moles of CO. This amount(ie 1150moles) is little compared to 1339.3moles of CO calculated from the question. Therefore, H2 is the limiting reactant.

Now we can calculate the theoretical yield as follows:

CO + 2H2 —> CH3OH

Molar Mass of H2 = 2g/mol

Mass of H2 from the balanced equation = 2 x 2 = 4g

Molar Mass of CH3OH = 12 + 3 + 16 + 1 = 32g/mol

From the equation,

4g of H2 produced 32g of CH3OH

Therefore, 4600g of H2 will produce = (4600 x 32)/4 = 36800g of CH3OH

Therefore, the theoretical yield is 36800g

B. Actual yield = 1.83 x 10^4g

theoretical yield = 36800g

%yield =?

%yield = Actual yield /Theoretical yield x 100

%yield = 1.83 x 10^4/36800 x 100

%yield = 49.73%

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