Calculate the mass of water produced when 6.51 g of butane reacts with excess oxygen.
Well, let's do some math and find out!
First, we need to figure out the balanced chemical equation for the reaction between butane and oxygen.
C₄H₁₀ + O₂ → CO₂ + H₂O
From the equation, we can see that for every 1 mole of butane reacting, we get 4 moles of water. Let's convert 6.51 g of butane to moles.
The molar mass of butane (C₄H₁₀) is about 58.12 g/mol, so:
6.51 g / 58.12 g/mol = 0.112 mol of butane
Since the reaction has an excess of oxygen, we can assume that all the butane reacts. Therefore, we can conclude that 0.112 mol of butane will produce 0.112 mol × 4 mol/1 mol = 0.448 mol of water.
Lastly, let's convert the moles of water to grams. The molar mass of water (H₂O) is about 18.02 g/mol, so:
0.448 mol × 18.02 g/mol = 8.08 g of water
Therefore, when 6.51 g of butane reacts with excess oxygen, the mass of water produced is approximately 8.08 g.
To calculate the mass of water produced, we need to know the balanced chemical equation for the reaction of butane with oxygen.
The balanced equation for the combustion of butane is:
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)
From the equation, we can see that for every 2 moles of butane (C4H10) that react, we obtain 10 moles of water (H2O) as a product.
To find the moles of butane in 6.51 g, we can use the molar mass of butane (C4H10), which is 58.12 g/mol.
Moles of butane = mass of butane / molar mass of butane
Moles of butane = 6.51 g / 58.12 g/mol
Moles of butane = 0.112 mol
Since the reaction is with an excess of oxygen, we can assume all the butane reacts.
Now, using the mole ratio between butane and water, we can calculate the moles of water produced.
Moles of water = Moles of butane x (10 moles of water / 2 moles of butane)
Moles of water = 0.112 mol x (10/2)
Moles of water = 0.56 mol
Finally, we can calculate the mass of water produced using the molar mass of water (H2O), which is 18.02 g/mol.
Mass of water = Moles of water x molar mass of water
Mass of water = 0.56 mol x 18.02 g/mol
Mass of water = 10.11 g
Therefore, the mass of water produced when 6.51 g of butane reacts with excess oxygen is 10.11 g.
To calculate the mass of water produced, you need to understand the balanced chemical equation for the reaction between butane and oxygen:
C4H10 + 13/2O2 --> 4CO2 + 5H2O
From the balanced equation, we can see that for every 1 mole of butane (C4H10) that reacts, 5 moles of water (H2O) are produced.
First, calculate the number of moles of butane:
Molar mass of butane (C4H10) = (4 x Atomic mass of carbon) + (10 x Atomic mass of hydrogen)
= (4 x 12.01 g/mol) + (10 x 1.01 g/mol)
= 48.04 g/mol + 10.10 g/mol
= 58.14 g/mol
Moles of butane = Mass of butane / Molar mass of butane
= 6.51 g / 58.14 g/mol
= 0.112 moles
Since the reaction is said to occur with excess oxygen, we assume all the moles of butane react.
Now that we know the number of moles of butane, we can calculate the moles of water produced:
Moles of water = Moles of butane x Coefficient ratio
= 0.112 moles x (5 moles of water / 1 mole of butane)
= 0.560 moles
Finally, calculate the mass of water produced:
Mass of water = Moles of water x Molar mass of water
= 0.560 moles x (2 x Atomic mass of hydrogen + Atomic mass of oxygen)
= 0.560 moles x (2 x 1.01 g/mol + 16.00 g/mol)
= 0.560 moles x (2.02 g/mol + 16.00 g/mol)
= 0.560 moles x 18.02 g/mol
= 10.09 g
Therefore, the mass of water produced when 6.51 g of butane reacts with excess oxygen is approximately 10.09 g.
2C4H10 + 13O2 ==> 8CO2 + 10H2O
moles C4H10 = grams/molar mass = 6.51/58 = 0.112
mols H2O produced = 0.112 mols C4H10 x (10 moles H2O/2 mols C4H10) =
0.112 x 5 = 0.56
g H2O = mols x molar mass = ?