Calculate the mass of water produced when 9.53
g
of butane reacts with excess oxygen
2C4H10 + 13O2 ==> 8CO2 + 10H2O
mols C4H10 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols C4H10 to mols H2O.
Now convert mols H2O to grams H2O with g = mols x molar mass = ?
To calculate the mass of water produced when butane reacts with excess oxygen, we need to first balance the chemical equation for the combustion reaction of butane, C4H10.
C4H10 + 13/2 O2 -> 4 CO2 + 5 H2O
From the balanced equation, we can see that for every 1 mole of butane (C4H10) reacted, we obtain 5 moles of water (H2O).
To find the moles of butane, we need to divide the given mass of butane by its molar mass.
Molar mass of C4H10: (4 * 12.01 g/mol) + (10 * 1.01 g/mol) = 58.12 g/mol
Moles of butane = 9.53 g / 58.12 g/mol ≈ 0.164 moles
Since we have a 1:5 ratio between butane and water, we can calculate the moles of water produced.
Moles of water = Moles of butane * 5 = 0.164 moles * 5 = 0.82 moles
Finally, to find the mass of water, we multiply the moles of water by its molar mass.
Molar mass of H2O: 2 * 1.01 g/mol + 16.00 g/mol = 18.02 g/mol
Mass of water = Moles of water * Molar mass of H2O = 0.82 moles * 18.02 g/mol ≈ 14.76 g
Therefore, approximately 14.76 grams of water will be produced when 9.53 grams of butane reacts with excess oxygen.
To solve this problem, we need to know the balanced chemical equation for the reaction between butane and oxygen.
C4H10 + O2 → CO2 + H2O
From the balanced equation, we can see that for every mole of butane (C4H10), we get 4 moles of water (H2O). Now, let's calculate the number of moles of butane using its molar mass.
Molar mass of butane (C4H10) = (4 * 12.01 g/mol) + (10 * 1.01 g/mol) = 58.12 g/mol
Number of moles of butane = Mass of butane / Molar mass of butane
= 9.53 g / 58.12 g/mol
≈ 0.164 moles
Since the reaction is given with excess oxygen, it means all the butane will react completely. So, the number of moles of water produced will be 4 times the number of moles of butane.
Number of moles of water = 4 * Number of moles of butane
= 4 * 0.164 mol
= 0.656 mol
Finally, let's calculate the mass of water produced.
Mass of water = Number of moles of water * Molar mass of water
= 0.656 mol * 18.02 g/mol
≈ 11.8 g
Therefore, when 9.53 g of butane reacts with excess oxygen, approximately 11.8 g of water is produced.