2so2(g) + o2(g) → 2so3(g) how many grams of so3 will be produced from the reaction of 6.60 liters o2 with excess so2 at stp? a) 47.3 g b) 64.1 g c) 108 g d) 543 g

A) 47.3 g

Explanation:

2SO2(g) + O2(g) → 2SO3(g)

From the balanced equation above, we can tell that;

1 mol of oxygen molecule produces 2 mol of SO3 molecule.

At Standard Temperature and Pressure (STP), 1 mole of any gas will occupy a volume of 22.4 L.

This means 22.4 L of O2 produces 2 mol of SO3.

In order to calculate the mass of SO3 produced, we use the formular;

mass = Molar mass * Number of moles

Molar mass of SO3 = 32 + 16*3 = 32 + 48 = 80 g/mol

mass = 80 * 2 = 160g

This meas 22.4L produces 160g. How many g would 6.60 L produce?

22.4 = 160

6.60 = x

Upon cross multiplication,

x = (6.60 * 160) / 22.4

x = 47.14g