I can't get this!
If 4.00 g of H2 are made to react with excess CO(g) how many grams of CH3OH(l) can theoretically be produced according to the following equation?
CO(g) + 2H2(g) → CH3OH(l)
This is a simple stoichiometry problem. Here are the steps. You need to learn how to solve them. Remember the steps.This will work 75% of the problems you have in beginning chemistry. With a few modifications it will work many more than that.
Step 1. Write and balance the equation. You've done that.
Step 2. Convert what you have into mols. mols H2 = grams/molar mass = 4/2 = 2
Step 3. Using the coefficients in the balanced equation, convert mols of what you have to mols of what you want.
2 mols H2 x (1 mol CH3OH/2 mols H2) = 1 mol CH3OH
Step 4. Now convert to grams. grams CH3OH = mols CH3OH x molar mass CH3OH = 1 x 32 = 32 g
REMEMBER the steps for doing this. It will save you a lot of heartache down the road.
Well, well, well, looks like we have a reaction on our hands! Let's do some math magic!
First, let's find out how many moles of H2 we have. We know we have 4.00 grams, but we need to convert that into moles. The molar mass of H2 is about 2 g/mol, so:
4.00 g H2 / 2 g/mol = 2.00 moles of H2
Remember, we have an excess of CO, so we're only concerned about the limiting reagent, which is H2 in this case. According to the balanced equation, 1 mole of CO produces 1 mole of CH3OH. Since we have 2.00 moles of H2, we can safely say that:
2.00 moles of H2 = 2.00 moles of CH3OH
To get the grams of CH3OH, we need to multiply the moles by its molar mass. The molar mass of CH3OH is approximately 32 g/mol, so:
2.00 moles CH3OH x 32 g/mol = 64.00 grams of CH3OH
So, there you have it! Theoretically, you can produce 64.00 grams of CH3OH. Just remember that in reality, reaction yields might not always be perfect.
To find out how many grams of CH3OH(l) can be produced, we need to use stoichiometry. The balanced equation tells us the ratio of the reactants and products.
The molar masses are:
H2: 2.02 g/mol
CO: 28.01 g/mol
CH3OH: 32.04 g/mol
First, we need to find out how many moles of H2 we have:
4.00 g H2 * (1 mol H2 / 2.02 g H2) = 1.98 mol H2
According to the balanced equation, the ratio of moles of CO to moles of H2 is 1:2. So, the moles of CO reacting with H2 will be twice the number of moles of H2.
Moles of CO = 2 * moles of H2 = 2 * 1.98 mol H2 = 3.96 mol CO
Now, we can use the stoichiometry to find the number of moles of CH3OH produced.
According to the balanced equation, the ratio of moles of CH3OH to moles of CO is 1:1.
So, the moles of CH3OH = 3.96 mol CO
Finally, we can find the mass of CH3OH produced.
Mass of CH3OH = moles of CH3OH * molar mass of CH3OH
Mass of CH3OH = 3.96 mol * 32.04 g/mol = 126.88 g
Therefore, theoretically, 126.88 grams of CH3OH can be produced.
To determine the mass of CH3OH(l) that can be produced, we need to use stoichiometry - the molar ratio of the reactants and products.
First, we need to find the number of moles of H2 involved in the reaction. We can use the molar mass of H2 to convert the given mass (4.00 g) to moles. The molar mass of H2 is 2.02 g/mol, so:
Number of moles of H2 = mass of H2 / molar mass of H2
= 4.00 g / 2.02 g/mol
≈ 1.98 mol
According to the balanced equation, for every 2 moles of H2, 1 mole of CH3OH is produced. So, using the mole ratio, we can find the number of moles of CH3OH that can be produced.
Number of moles of CH3OH = (Number of moles of H2) / 2
≈ 1.98 mol / 2
≈ 0.99 mol
Finally, we can find the mass of CH3OH produced by multiplying the number of moles of CH3OH by its molar mass. The molar mass of CH3OH is 32.04 g/mol.
Mass of CH3OH = Number of moles of CH3OH × molar mass of CH3OH
≈ 0.99 mol × 32.04 g/mol
≈ 31.72 g
Therefore, the theoretical yield of CH3OH(l) is approximately 31.72 grams.