Question

Use the References to access Important values If needed for this question. For the following reaction, 20.6 grams of sulfur dioxide are allowed to react with 9.77 grams of oxygen gas. sulfur dioxide (g) + oxygen (g) → sulfur trioxide (g) What is the maximum amount of sulfur trioxide that can be formed? What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 7 more group attempts remaining

Answer 1

1)

Molar mass of SO2,

MM = 1*MM(S) + 2*MM(O)

= 1*32.07 + 2*16.0

= 64.07 g/mol

mass(SO2)= 20.6 g

use:

number of mol of SO2,

n = mass of SO2/molar mass of SO2

=(20.6 g)/(64.07 g/mol)

= 0.3215 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 9.77 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(9.77 g)/(32 g/mol)

= 0.3053 mol

Balanced chemical equation is:

2 SO2 + O2 ---> 2 SO3 +

2 mol of SO2 reacts with 1 mol of O2

for 0.3215 mol of SO2, 0.1608 mol of O2 is required

But we have 0.3053 mol of O2

so, SO2 is limiting reagent

we will use SO2 in further calculation

Molar mass of SO3,

MM = 1*MM(S) + 3*MM(O)

= 1*32.07 + 3*16.0

= 80.07 g/mol

According to balanced equation

mol of SO3 formed = (2/2)* moles of SO2

= (2/2)*0.3215

= 0.3215 mol

use:

mass of SO3 = number of mol * molar mass

= 0.3215*80.07

= 25.74 g

Answer: 25.7 g

2)

SO2 is limiting reagent

Answer: SO2

3)

According to balanced equation

mol of O2 reacted = (1/2)* moles of SO2

= (1/2)*0.3215

= 0.1608 mol

mol of O2 remaining = mol initially present - mol reacted

mol of O2 remaining = 0.3053 - 0.1608

mol of O2 remaining = 0.1446 mol

Molar mass of O2 = 32 g/mol

use:

mass of O2,

m = number of mol * molar mass

= 0.1446 mol * 32 g/mol

= 4.626 g

Answer: 4.63 g