What is the theoretical yield of SO produced by 5.99 g of S?
To determine the theoretical yield of SO2 produced by 5.99 g of S, we need to first write a balanced chemical equation for the reaction between S and O2 to form SO2.
The balanced chemical equation is:
S + O2 -> SO2
From the balanced chemical equation, we can see that the molar ratio between S and SO2 is 1:1.
Next, we need to convert the mass of S to moles. The molar mass of S is 32.07 g/mol.
5.99 g of S x (1 mol S / 32.07 g S) = 0.187 mol S
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 5.99 g of S is 0.187 mol.
Finally, we can convert the number of moles of SO2 to grams using the molar mass of SO2 (64.07 g/mol).
0.187 mol SO2 x 64.07 g SO2 / 1 mol SO2 = 11.98 g SO2
Therefore, the theoretical yield of SO2 produced by 5.99 g of S is 11.98 grams.