calculate the mass carbon monoxide, CO, that was needed to produce 78g of methanol, CH3OH, by the following reaction: 2H2 + CO - CH3OH.
sorry, typed incorrectly...will repost.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the mass of carbon monoxide (CO) needed to produce 78g of methanol (CH3OH) using the given reaction:
2H2 + CO -> CH3OH
We first need to determine the molar mass of methanol (CH3OH) and carbon monoxide (CO).
The molar mass of methanol (CH3OH) can be calculated as follows:
Molar mass (CH3OH) = (C: 1 x 12.01 g/mol) + (H: 4 x 1.01 g/mol) + (O: 1 x 16.00 g/mol)
= 12.01 g/mol + 4.04 g/mol + 16.00 g/mol
= 32.05 g/mol
Now, we can calculate the moles of methanol produced using the given mass:
Moles of CH3OH = Mass of CH3OH / Molar mass of CH3OH
= 78g / 32.05 g/mol
= 2.433 mol
From the balanced equation, we know that 1 mole of CO is required to produce 1 mole of CH3OH. Therefore, the moles of CO required would be the same as the moles of CH3OH.
Moles of CO = Moles of CH3OH
= 2.433 mol
Finally, we can calculate the mass of carbon monoxide (CO) using its molar mass:
Mass of CO = Moles of CO x Molar mass of CO
= 2.433 mol x 28.01 g/mol
= 68.11 g
Therefore, approximately 68.11 grams of carbon monoxide (CO) is needed to produce 78 grams of methanol (CH3OH) in the given reaction.
To calculate the mass of carbon monoxide (CO) needed to produce 78g of methanol (CH3OH) through the given reaction: 2H2 + CO -> CH3OH, we need to use the concept of stoichiometry.
First, let's calculate the molar mass of CH3OH (methanol):
C = 12.01 g/mol
H = 3.01 g/mol x 4 = 12.04 g/mol
O = 16.00 g/mol + 1.01 g/mol = 17.01 g/mol
Molar mass of CH3OH = 12.01 g/mol + 12.04 g/mol + 17.01 g/mol = 41.06 g/mol
Next, we need to calculate the moles of CH3OH using its molar mass and the given mass:
Moles of CH3OH = Mass of CH3OH / Molar mass of CH3OH
Moles of CH3OH = 78g / 41.06 g/mol ≈ 1.898 mol
From the balanced equation, we can see that 1 mol of CO is needed to produce 1 mol of CH3OH. Therefore, the moles of CO required will also be 1.898 mol.
Finally, to calculate the mass of CO required, we need to use its molar mass:
Molar mass of CO = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
Mass of CO = Moles of CO × Molar mass of CO
Mass of CO = 1.898 mol × 28.01 g/mol ≈ 53.06 g
Therefore, the mass of carbon monoxide (CO) needed to produce 78g of methanol (CH3OH) through the given reaction is approximately 53.06 grams.