gaseous butane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. Suppose 38.9 g of butane is mixed with 35. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 significant digits.
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write and balance the equation:
C4H10 + 13/2 O2 >>> 4CO2 + 5H2O
so for each mole of butane, you get fur moles five moles of water.
MolesButane=38.9/molmassbutane
= 38.9/58.2
moles water= 4* above
mass water= moleswater*18 grams
what is the proper formula of Gaseous Butane (C4H10) reacts with Gaseous Oxygen (O2) in a combustion reaction to form Carbon dioxide Gas and Gaseous Water (Dihydrogen monoxide)?
To calculate the maximum mass of water produced, we need to determine which reactant is limiting and then use stoichiometry to convert the limiting reactant into the product.
Step 1: Determine the limiting reactant.
To find the limiting reactant, we compare the number of moles of each reactant to the stoichiometric ratio given by the balanced equation.
The balanced equation for the reaction is:
2 C4H10(g) + 13 O2(g) -> 8 CO2(g) + 10 H2O(g)
Let's calculate the number of moles for each reactant:
Number of moles of butane (C4H10):
38.9 g C4H10 * (1 mol C4H10 / molar mass of C4H10) = 38.9 g C4H10 * (1 mol C4H10 / 58.12 g C4H10) = 0.6699 mol C4H10
Number of moles of oxygen (O2):
35.0 g O2 * (1 mol O2 / molar mass of O2) = 35.0 g O2 * (1 mol O2 / 32.00 g O2) = 1.09375 mol O2
According to the balanced equation, the stoichiometric ratio between butane and oxygen is 2 mol C4H10 : 13 mol O2. Therefore, we need to compare the moles of each reactant.
The ratio of moles of C4H10 to moles of O2 is:
0.6699 mol C4H10 / 2 mol C4H10 = 0.3349
1.09375 mol O2 / 13 mol O2 = 0.0841
From the molar ratio, we can see that the ratio of moles of butane to the moles of oxygen is approximately 0.3349:0.0841 or 4:1.
Since the ratio of the moles of butane to oxygen is 4:1, it means we have excess oxygen. Therefore, the limiting reactant is oxygen.
Step 2: Calculate the maximum mass of water produced.
To calculate the maximum mass of water produced, we need to use the stoichiometric ratio between oxygen and water.
From the balanced equation, the stoichiometric ratio between oxygen and water is 13 mol O2 : 10 mol H2O.
Number of moles of water (H2O):
Using stoichiometry,
1.09375 mol O2 * (10 mol H2O / 13 mol O2) = 0.84295 mol H2O
Now, we can convert the moles of water into grams:
Mass of water:
0.84295 mol H2O * (molar mass of H2O / 1 mol H2O) = 0.84295 mol H2O * (18.02 g / 1 mol) = 15.18 g H2O
Therefore, the maximum mass of water that could be produced by the chemical reaction is 15.18 grams.