What mass of sulfur must be used to produce 12.6L of sulfur dioxide (SO2) at STP?
S8 + O2 SO2
moles SO2=12.6/24.0 moles
mass S8=molemassS8*12.6/24
OOOPs, you didn't balance the equation.
S8 +8 O2 >>> 8SO2
moles SO2=12.6/24
moles S8= 1/8*12.6/24
grams S8= moles S8*molmassS8
To find the mass of sulfur needed to produce 12.6 L of sulfur dioxide (SO2) at standard temperature and pressure (STP), we need to use the given balanced chemical equation.
The balanced equation shows that 1 mole of S8 reacts with 1 mole of O2 to produce 8 moles of SO2.
First, we need to find the moles of sulfur dioxide (SO2). The volume of gas at STP (Standard Temperature and Pressure) is directly proportional to the number of moles of the gas. At STP, 1 mole of an ideal gas occupies 22.4 liters.
Therefore, we can calculate the moles of SO2 as follows:
moles of SO2 = volume of SO2 / molar volume at STP
moles of SO2 = 12.6 L / 22.4 L/mol
moles of SO2 = 0.5625 mol
Since the stoichiometry of the balanced equation is 1:8 (1 mole of S8 reacts to produce 8 moles of SO2), we can find the moles of S8 as follows:
moles of S8 = moles of SO2 / stoichiometric ratio
moles of S8 = 0.5625 mol / 8
moles of S8 = 0.07031 mol
Now, we need to find the molar mass of sulfur (S) to determine the mass of sulfur:
The molar mass of S = 32.07 g/mol
mass of S = moles of S8 x molar mass of S
mass of S = 0.07031 mol x 32.07 g/mol
mass of S = 2.25 g
Therefore, approximately 2.25 grams of sulfur must be used to produce 12.6 L of sulfur dioxide at STP.