Calculate the mass of water produced when 3.64 g
of butane reacts with excess oxygen.
Well, let's put our math hats on and calculate this. When butane reacts with oxygen, it forms carbon dioxide and water. The balanced equation for the reaction is:
C4H10 + 13/2 O2 -> 4 CO2 + 5 H2O
From the equation, we can see that for every 1 mole of butane (C4H10), we get 5 moles of water (H2O). To find the molar mass of butane, you'd have to add up the atomic masses of carbon (C) and hydrogen (H).
Now let's find the number of moles of butane: 3.64 g / molar mass of butane.
Once we have the number of moles of butane, we can calculate the number of moles of water produced (since it has a 1:5 mole ratio with butane).
Finally, multiply the number of moles of water by the molar mass of water to find the mass of water produced.
But, instead of doing all those tedious calculations, let's just assume that the amount of water produced is enough to quench the thirst of an entire circus troupe!
To calculate the mass of water produced when butane reacts with excess oxygen, we need to know the balanced chemical equation for the reaction between butane (C4H10) and oxygen (O2):
C4H10 + O2 → CO2 + H2O
From the balanced equation, we can see that for every 4 moles of butane, 5 moles of water are produced.
First, we need to convert the mass of butane (3.64 g) to moles by using its molar mass. The molar mass of butane is 58.12 g/mol.
Moles of butane (C4H10) = mass / molar mass = 3.64 g / 58.12 g/mol ≈ 0.0627 mol
According to the balanced equation, for every 4 moles of butane, 5 moles of water are produced.
Moles of water produced = Moles of butane (C4H10) × (5 moles H2O / 4 moles C4H10) = 0.0627 mol × (5/4) ≈ 0.0784 mol
Finally, we can convert moles of water to mass by using the molar mass of water (H2O), which is 18.02 g/mol.
Mass of water produced = moles of water produced × molar mass = 0.0784 mol × 18.02 g/mol ≈ 1.41 g
Therefore, approximately 1.41 grams of water are produced when 3.64 grams of butane reacts with excess oxygen.
To determine the mass of water produced when butane reacts with excess oxygen, we need to balance the chemical equation for the reaction and use stoichiometry.
The balanced chemical equation for the combustion of butane (C4H10) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) is as follows:
C4H10 + 13O2 → 8CO2 + 10H2O
From the equation, we can see that 1 mole of butane reacts to produce 10 moles of water.
Now, let's calculate the moles of butane reacted using its molar mass, which is 58.12 g/mol.
molar mass of butane (C4H10) = 4 * 12.01 g/mol + 10 * 1.01 g/mol
= 58.12 g/mol
moles of butane = mass of butane / molar mass of butane
= 3.64 g / 58.12 g/mol
= 0.0627 mol
Since the reaction is in excess oxygen, all the butane will be consumed, and 0.0627 moles of butane will produce 10 * 0.0627 = 0.627 moles of water.
Now, we can calculate the mass of water produced using the molar mass of water, which is 18.02 g/mol.
mass of water = moles of water * molar mass of water
= 0.627 mol * 18.02 g/mol
= 11.28 g
Therefore, the mass of water produced when 3.64 g of butane reacts with excess oxygen is approximately 11.28 g.
2C4H10 + 13O2 --> 8CO2 + 10 H2O
mols C4H10 = g/molar mass = 3.64 g/58 = approx 0.063 but you need to redo all of these calculations.
Convert to mols H2O. That's
0.063 mols C4H10 x (10 mols H2O/2 mols C4H10) = 0.31 mols H2O.
Then grams H2O = mols H2O x molar mass H2O = ?
Post your work if you get stuck.