Calculate the mass of water produced when 4.36g of butane reacts with excess oxygen.
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To calculate the mass of water produced when butane reacts with excess oxygen, we need to balance the chemical equation for the reaction first:
C4H10 + O2 → CO2 + H2O
From the balanced equation, we can see that 1 mole of C4H10 reacts to produce 4 moles of H2O. We can use this ratio to calculate the mass of water produced.
1 mole of C4H10 has a molar mass of 58.12 g/mol. To find the number of moles of butane, we divide the given mass by its molar mass:
moles of C4H10 = 4.36 g / 58.12 g/mol ≈ 0.075 moles
Since we have an excess of oxygen, all the butane will react completely. Therefore, 0.075 moles of C4H10 will produce 0.075 moles of H2O.
The molar mass of H2O is 18.02 g/mol. To find the mass of water produced, we multiply the number of moles by the molar mass:
mass of H2O = 0.075 moles * 18.02 g/mol ≈ 1.35 g
Therefore, the mass of water produced when 4.36 g of butane reacts with excess oxygen is approximately 1.35 g.
To calculate the mass of water produced when 4.36g of butane reacts with excess oxygen, we need to know the balanced chemical equation for the reaction between butane (C4H10) and oxygen (O2).
The balanced equation for this combustion reaction is:
2C4H10 + 13O2 -> 8CO2 + 10H2O
From the balanced equation, we can see that for every 2 moles of butane, 10 moles of water are produced.
To calculate the mass of water produced, we need to determine the number of moles of butane and convert it to moles of water, and then finally convert moles of water to grams.
First, calculate the number of moles of butane:
moles of butane = mass of butane (g) / molar mass of butane (g/mol)
The molar mass of butane (C4H10) can be calculated by adding up the atomic masses of its constituent elements:
C = 12.01 g/mol (4 atoms)
H = 1.01 g/mol (10 atoms)
Molar mass of butane = (4 × 12.01 g/mol) + (10 × 1.01 g/mol) = 58.12 g/mol
Thus, the number of moles of butane = 4.36 g / 58.12 g/mol = 0.075 moles.
Since the balanced equation shows that 2 moles of butane react to produce 10 moles of water, we can set up a mole ratio:
0.075 moles butane × (10 moles water / 2 moles butane) = 0.375 moles water.
Finally, we calculate the mass of water produced:
mass of water (g) = moles of water × molar mass of water (g/mol)
The molar mass of water (H2O) is:
H = 1.01 g/mol (2 atoms)
O = 16.00 g/mol (1 atom)
Molar mass of water = (2 × 1.01 g/mol) + (1 × 16.00 g/mol) = 18.02 g/mol
Therefore, the mass of water produced = 0.375 moles × 18.02 g/mol = 6.76 grams.
So, when 4.36g of butane reacts with excess oxygen, 6.76g of water is produced.