Respuesta :
Answer:
See explanation below
Explanation:
Let's write and balance the reaction first:
C₄H₁₀ + 13/2O₂ -----> 4CO₂ + 5H₂O
Now that the reaction is balanced, we know that the water produced was 32.2 g. So, let's calculate the theorical yield of water using 42.24 g of butane and 62.3 g of oxygen.
The molar mass of butane is 58 g/mol, while water is 18 g/mol and oxygen 32 g/mol
The moles of each reactant are:
moles butane = 42.24 / 58 = 0.73 moles
moles oxygen = 62.3 / 32 = 1.95 moles
Now, let's calculate the limiting reactant:
1 mole butane -------> 13/2 mole oxygen
0.73 moles ---------> X
X = 0.73 * 13/2 = 4.75 moles oxygen
However, we only have 1.95 moles, therefore, oxygen is the limiting reactant.
As oxygen is the limiting reactant, this means that the moles produced of water will be:
moles water = 1.95 moles
Finally the mass of water:
m = 1.5 * 18 =¨35.1 g
The percent yield will be:
% = (32.2 / 35.1) * 100
% = 91.73 %
