Calculate the mass of water produced when 1.47 g of butane reacts with excess oxygen.
2C4H10 + 13 O2 ==> 8CO2 + 10H2O
mols C4H10 = grams/molar mass = 1.47/58 = ?
mols H2O produced = ? from previous step x (10 mols H2O/2 mols C4H10) = ? from previous step x (10/2) = xx
Then convert to grams. grams = xx mols H2O x molar mass H2O
Post your work if you get stuck.
ecdwjnc
To calculate the mass of water produced when 1.47 g of butane reacts with excess oxygen, we need to understand the balanced chemical equation for the reaction.
The balanced equation for the combustion of butane can be represented as:
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
From the equation, we can see that 2 moles of butane (C4H10) react with 13 moles of oxygen (O2) to produce 10 moles of water (H2O).
First, let's calculate the number of moles of butane:
1.47 g of butane is equal to:
(1.47 g) / (58.12 g/mol) = 0.0253 moles
Now, let's determine the number of moles of water produced. Since the stoichiometric coefficient of water in the balanced equation is 10, we can calculate:
Number of moles of water produced = 10 × Number of moles of butane
= 10 × 0.0253 moles
= 0.253 moles
Finally, let's convert the number of moles of water to mass using the molar mass of water:
Mass of water produced = (0.253 moles) × (18.02 g/mol)
= 4.56 g
Therefore, when 1.47 g of butane reacts with excess oxygen, it produces approximately 4.56 g of water.
To calculate the mass of water produced when 1.47 g of butane reacts with excess oxygen, we need to use stoichiometry and the balanced chemical equation for the reaction. In this case, the balanced equation for the combustion of butane is:
2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O
From the balanced equation, we can see that for every 2 moles of butane, we produce 10 moles of water. Therefore, we need to convert the mass of butane (1.47 g) to moles, and then use the stoichiometric ratio to calculate the mass of water.
Step 1: Calculate the moles of butane:
To do this, we need to know the molar mass of butane (C4H10). The molar mass of C4H10 is:
(4*12.01 g/mol) + (10*1.01 g/mol) = 58.14 g/mol
Moles of butane = Mass of butane / Molar mass of butane
Moles of butane = 1.47 g / 58.14 g/mol
Step 2: Calculate the moles of water:
Using the stoichiometric ratio, we know that for every 2 moles of butane, we produce 10 moles of water.
Moles of water = Moles of butane * (10 moles of water / 2 moles of butane)
Step 3: Calculate the mass of water:
To calculate the mass of water, we need to know the molar mass of water (H2O), which is:
(2*1.01 g/mol) + (1*16.00 g/mol) = 18.02 g/mol
Mass of water = Moles of water * Molar mass of water
Now, we can plug in the values to calculate the mass of water:
Moles of butane = 1.47 g / 58.14 g/mol
Moles of water = (1.47 g / 58.14 g/mol) * (10 mol/2 mol)
Mass of water = (1.47 g / 58.14 g/mol) * (10 mol/2 mol) * 18.02 g/mol
Calculating this expression will give us the mass of water produced when 1.47 g of butane reacts with excess oxygen.