What is the theoretical yield of SO produced by 6.16g of S?
To calculate the theoretical yield of SO2 produced by 6.16g of S, we need to first determine the molar ratio between S and SO2 in the balanced chemical equation.
The balanced chemical equation for the reaction between S and O2 to produce SO2 is:
S + O2 -> SO2
From the equation, we can see that the molar ratio between S and SO2 is 1:1.
Next, we need to convert the mass of S (6.16g) to moles. The molar mass of S is 32.06 g/mol.
Number of moles of S = mass of S / molar mass of S
Number of moles of S = 6.16g / 32.06 g/mol
Number of moles of S = 0.1921 mol
Since the molar ratio between S and SO2 is 1:1, the number of moles of SO2 produced will be the same as the number of moles of S.
Therefore, the theoretical yield of SO2 produced by 6.16g of S is 0.1921 mol.
Now, we need to convert this number of moles of SO2 to grams. The molar mass of SO2 is 64.06 g/mol.
Theoretical yield of SO2 = number of moles of SO2 x molar mass of SO2
Theoretical yield of SO2 = 0.1921 mol x 64.06 g/mol
Theoretical yield of SO2 = 12.31 g
Therefore, the theoretical yield of SO2 produced by 6.16g of S is 12.31g.