What mass of methanol (Ch3OH) is produced when 86.04 of carbon monoxide reacts with 14.14g of hydrogen HELPP PLS!?!

98.33 g of CH3OH

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

CO(g) + 2H2(g) —> CH3OH(l)

Step 2:

Determination of the masses of CO and H2 that reacted and the mass of CH3OH produced from the balanced equation.

This can be obtained as follow:

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

Mass of CO from the balanced equation = 1 x 28 = 28 g

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

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

Molar mass of CH3OH = 12 + (3x1) + 16 + 1 = 32 g/mol

Mass of CH3OH from the balanced equation = 1 x 32 = 32 g.

Summary:

From the balanced equation above,

28 g of CO reacted with 4 g of H2 to produce 32 g of CH3OH.

Step 3:

Determination of the limiting reactant.

This can be obtained as follow:

From the balanced equation above,

28 g of CO reacted with 4 g of H2.

Therefore, 86.04 g of CO will react with = (86.04 x 4)/28 = 12.29 g of H2.

From the calculation made above, only 12.29 g out of 14.14 g of H2 given were required to react completely with 86.04 g of CO.

Therefore, CO is the limiting reactant and H2 is the excess reactant.

Step 4:

Determination of the mass of methanol, CH3OH produced from the reaction.

In this case the limiting reactant shall be used because it will give the maximum yield of methanol, CH3OH since all of it were consumed in the reaction.

The limiting reactant is CO and the mass of methanol, CH3OH produced can be obtained as follow:

From the balanced equation above,

28 g of CO reacted to produce 32 g of CH3OH.

Therefore, 86.04 g of CO will react to produce = (86.04 x 32)/28 = 98.33 g

Therefore, 98.33 g of CH3OH were obtained from the reaction.