calculate the mass carbon monoxide, CO, that was needed to produce 78g of methanol, CH3OH, by the following reaction: 2H2 + CO -> CH3OH.

answer the question someone!

Answer: loading

To calculate the mass of carbon monoxide (CO) required to produce 78g of methanol (CH3OH), we first need to determine the balanced equation for the reaction.

The given balanced equation is: 2H2 + CO -> CH3OH

From the equation, we can see that one molecule of methanol (CH3OH) is formed from one molecule of carbon monoxide (CO). This means that the molar ratio of CO to CH3OH is 1:1.

To find the amount of CO required, we need to convert the mass of CH3OH to moles, using its molar mass.

The molar mass of CH3OH is calculated as follows:
C = 12.01 g/mol
H = 1.01 g/mol (3 Hydrogen atoms)
O = 16.00 g/mol
Total molar mass = (12.01 + (1.01 x 3) + 16.00) g/mol = 32.04 g/mol

Now, we can calculate the number of moles of CH3OH:
Moles of CH3OH = Mass of CH3OH / Molar mass of CH3OH
Moles of CH3OH = 78 g / 32.04 g/mol = 2.433 mol

Since the molar ratio of CO to CH3OH is 1:1, we need an equal number of moles of CO to produce the given amount of CH3OH. Therefore, the number of moles of CO required is also 2.433 mol.

Finally, we can calculate the mass of CO required using its molar mass:

Molar mass of CO = (12.01 + 16.00) g/mol = 28.01 g/mol
Mass of CO = Moles of CO x Molar mass of CO
Mass of CO = 2.433 mol x 28.01 g/mol = 67.98 g

Therefore, approximately 68g of carbon monoxide (CO) is required to produce 78g of methanol (CH3OH) in the given reaction.

To calculate the mass of carbon monoxide (CO) needed to produce 78g of methanol (CH3OH) using the given reaction, we need to follow these steps:

Step 1: Find the molar mass of methanol (CH3OH)
The molar mass of CH3OH can be calculated by adding up the atomic masses of all its constituent atoms:
Carbon (C) has a molar mass of 12.01 g/mol
Hydrogen (H) has a molar mass of 1.01 g/mol

Molar mass of CH3OH = (1 * C) + (4 * H) + (1 * O) = (1 * 12.01) + (4 * 1.01) + (1 * 16.00) = 32.04 g/mol

Step 2: Determine the stoichiometric ratio between CO and CH3OH
From the balanced equation, we can see that 1 mole of CO reacts with 1 mole of CH3OH. Therefore, the stoichiometric ratio between CO and CH3OH is 1:1.

Step 3: Calculate the moles of CH3OH
To calculate the moles of CH3OH, divide the given mass (78g) by its molar mass:
Moles of CH3OH = mass / molar mass = 78g / 32.04 g/mol = 2.434 mol

Step 4: Calculate the moles of CO needed
Since the stoichiometry between CO and CH3OH is 1:1, the moles of CO needed will be equal to the moles of CH3OH.

Moles of CO = 2.434 mol

Step 5: Convert moles to mass
To convert moles to mass, we multiply the number of moles by the molar mass of CO:

Mass of CO = moles of CO * molar mass of CO = 2.434 mol * (12.01 g/mol + 16.00 g/mol) = 67.86 g

Therefore, the mass of carbon monoxide (CO) needed to produce 78g of methanol (CH3OH) is approximately 67.86 grams.