Chemistry Limiting Reactant

Question

For the following reaction, 20.4 grams of sulfur dioxide are allowed to react with 9.63 grams of oxygen gas .

sulfur dioxide(g) + oxygen(g)  sulfur trioxide(g)

What is the maximum mass of sulfur trioxide that can be formed?  

What is the FORMULA for the limiting reagent?

What mass of the excess reagent remains after the reaction is complete?  

Answer 1

2SO2+O2 --> 2SO3

20.4 g SO2

9.63 g O2

Mol SO2

20.4 ( 1mol / 64.058 g) = 0.3184613943 mol SO2

Mol O2

9.63 g (1mol / 32.00 g ) = 0.3009375 mol O2

Theoretical yield product from each substance:

0.3184613943 mol SO2 (2 mol SO3 / 2 mol SO2) = 0.3184613943 mol SO3

0.3009375 mol O2 (2 mol SO3 / 1 mol O2) = 0.601875 mol SO3

Smaller theoretical yield results from SO2, so SO2 is the limiting reactant (O2 is in excess), and the theoretical yield is the smaller value, 0.3184613943 mol SO3

Convert theoretical yield from mol to g with molar mass:

0.3184613943 mol SO3 (80.057 g SO3 / 1 mol SO3) = 25.495063 g SO3 theoretical yield (this is the maximum mass that can be formed)

What is the formula for the limiting reagent? SO2

What is the mass of excess reactant that remains?

Convert mol LR (SO2) to grams excess reactant (O2)

0.3184613943 mol SO2 (1 mol O2 / 2 mol SO2) = 0.15923069715 mol O2 reacts

Convert mol O2 to grams O2

0.15923069715 mol O2 reacts (32 g O2 / 1 mol O2)=5.0953823088 g O2 reacts

Excess mass O2 = initial mass O2 - mass O2 that reacts

Excess mass O2 = 9.63g O2 - 5.0953823088 g O2

Excess mass O2 = 4.53461769 g O2