WHEN 54.8g OF SILICON (iv) OXIDE IS HEATED WITH EXCESS OF CARBON, 32.5g OF SILICON CARBIDE IS PRODUCED BESIDS THE CARBON MONOXIDE GAS. WHAT IS THE PERCENTAGE YIELD OF THIS REACTION
SiO2 + 3C ==> SiC + 2CO
mols SiO2 = grams/molar mass = 54.8/60.08 = 0.912
From the coefficients in the equation you know 1 mol SiC is produced for every mole of SiO2; therefore, you will botain 0.912 mols SiC. Convert to grams. grams SiC = mols SiC x molar mass SiC = 0.912 x 40.09 = about 36.6 g SiC. That is the theoretical yield (TY). The actual yield (AY) is 32.5 from the problem.
% yield = (AY/TY)*100 = ?Check my arithmetic.
59.2%
To calculate the percentage yield of a reaction, you need to compare the actual yield to the theoretical yield.
The balanced chemical equation for the reaction is:
SiO2 + 3C -> SiC + 2CO
First, we need to calculate the theoretical yield of silicon carbide (SiC) using the given mass of silicon (IV) oxide (SiO2):
Molar mass of SiO2 = 60.08 g/mol
Molar mass of SiC = 40.1 g/mol
Convert the mass of SiO2 to moles by dividing by its molar mass:
54.8 g SiO2 * (1 mol SiO2 / 60.08 g SiO2) = 0.913 mol SiO2
According to the balanced equation, the mole ratio between SiO2 and SiC is 1:1. Therefore, the number of moles of SiC produced is also 0.913 mol.
Calculate the theoretical mass of SiC using its molar mass:
0.913 mol SiC * (40.1 g SiC / 1 mol SiC) = 36.61 g SiC
Now, we can calculate the percentage yield:
Percentage yield = (Actual yield / Theoretical yield) * 100
The actual yield is given as 32.5 g of SiC.
Percentage yield = (32.5 g / 36.61 g) * 100 = 88.92%
Therefore, the percentage yield of this reaction is approximately 88.92%.
To find the percentage yield of the reaction, you need to compare the actual yield (32.5g of silicon carbide) to the theoretical yield (the maximum amount of silicon carbide that could be produced based on the amount of silicon(IV) oxide used).
First, you need to determine the molecular formula of silicon carbide (also known as silicon(IV) carbide or SiC) and silicon(IV) oxide (also known as silica or SiO2). The molecular formula for silicon carbide is SiC, while the molecular formula for silicon oxide is SiO2.
Next, calculate the molar masses of silicon carbide and silicon oxide:
- The molar mass of SiC = atomic mass of silicon (28.09 g/mol) + atomic mass of carbon (12.01 g/mol) = 40.10 g/mol
- The molar mass of SiO2 = atomic mass of silicon (28.09 g/mol) + 2 x atomic mass of oxygen (16.00 g/mol) = 60.08 g/mol
Now, calculate the number of moles of silicon oxide used:
Number of moles = mass / molar mass
Number of moles of SiO2 = 54.8 g / 60.08 g/mol = 0.912 mol
According to the balanced chemical equation for the reaction between silicon oxide and carbon:
SiO2(s) + 3C(s) → SiC(s) + 2CO(g)
The stoichiometric ratio between silicon oxide and silicon carbide is 1:1. This means that for every 1 mole of silicon oxide, 1 mole of silicon carbide is produced.
Therefore, the theoretical yield of silicon carbide would also be 0.912 moles.
Next, calculate the mass of the theoretical yield of silicon carbide:
Mass = number of moles x molar mass
Mass of SiC (theoretical yield) = 0.912 mol x 40.10 g/mol = 36.48 g
Now, you can calculate the percentage yield of the reaction:
Percentage yield = (actual yield / theoretical yield) x 100%
Percentage yield = (32.5 g / 36.48 g) x 100%
Percentage yield ≈ 89.17%
Therefore, the percentage yield of this reaction is approximately 89.17%.