Methanol (CH3OH) is used as fuel in race cars and is a potential replacement for gasoline.  It is synthesized by combination of hydrogen and carbon monoxide.  Suppose 65.8 g of Carbon monoxide reacts with 8.6 g of hydrogen.  What is the theoretical yield in moles and grams?

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The synthesis of methanol from hydrogen and carbon monoxide takes place following the chemical reaction

2*H2 +CO = CH3OH

The molecular masses of the reactants and of the reaction product are

M(H2) =2 g

M(CO) =12 +16 =28 g

M(CH3OH) = 12+16 +4 = 32 g

Thus for 65.8 grams of CO corresponds 65.8/28 = 2.35 moles of CO

For 8.6 grams of H2 corresponds 8.6/2 = 4.3 moles of H2

Since in the reaction above for 1 mole of CO entering the reaction we need 2 moles of H2 as a by-reactant, for 2.35 moles of CO (65.8 grams) one needs 2*2.35 =4.7 moles of H2 to enter reaction (and we have only 4.3 moles H2). Therefore hydrogen is the limiting reactant here.

From the above reaction we can write the following mass balance

2*2 grams of H2 gives......32 grams CH3OH

8.6 grams of H2 gives ......x grams CH3OH

x =32*8.6/4 =68.8 grams CH3OH

The number of moles of CH3OH in 68.8 grams of substance is

n =68.8/32 =2.15 moles CH3OH

The theoretical yield of the reaction is 68.8 grams (2.15 moles) of CH3OH.

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