Calculate the mass of carbon monoxide that was needed to produce 78 grams of methanol by the following reaction:

2H2+CO=CH3OH

To calculate the mass of carbon monoxide (CO) needed to produce 78 grams of methanol (CH3OH) using the given reaction, we need to determine the stoichiometry of the reaction and use it to find the molar ratio between carbon monoxide and methanol.

Looking at the balanced equation:
2H2 + CO = CH3OH

The coefficient in front of carbon monoxide (CO) is 1, indicating that 1 mole of CO reacts with 1 mole of methanol.

Molar mass of CH3OH (methanol) = 12.01 g/mol (C) + 3(1.01 g/mol) (H) + 16.00 g/mol (O) = 32.04 g/mol

Now, we can set up a proportion to calculate the mass of CO needed.

(1 mole CO / 1 mole CH3OH) = (x g CO / 78 g CH3OH)

Cross-multiplying the equation gives us:

x g CO = (1 mole CO / 1 mole CH3OH) * (78 g CH3OH)

To find the mole ratio between CO and CH3OH, we need to calculate the molar quantity of CH3OH:

(78 g CH3OH) / (32.04 g/mol CH3OH) = 2.43 moles CH3OH

Now, we can calculate the mass of CO:

x g CO = (1 mole CO / 1 mole CH3OH) * (2.43 moles CH3OH) * (28.01 g/mol CO)

x g CO = 68.89 g CO

Therefore, 68.89 grams of carbon monoxide (CO) were needed to produce 78 grams of methanol (CH3OH) in the given reaction.