Question

Constants | Periodic Table Iron(III) oxide reacts with carbon to give iron and carbon monoxide. Fe2O3(s)+3C(s)→2Fe(s)+3CO(g)

Part A How many grams of C are required to react with 78.2 g of Fe2O3? nothing   g   SubmitRequest Answer Part B How many grams of CO are produced when 38.0 g of C reacts? nothing   g   SubmitRequest Answer Part C How many grams of Fe can be produced when 6.40 g of Fe2O3 reacts? nothing   g   SubmitRequest Answer

Answer 1

A)

Molar mass of Fe2O3,

MM = 2*MM(Fe) + 3*MM(O)

= 2*55.85 + 3*16.0

= 159.7 g/mol

mass of Fe2O3 = 78.2 g

mol of Fe2O3 = (mass)/(molar mass)

= 78.2/1.597*10^2

= 0.4897 mol

According to balanced equation

mol of C required = (3/1)* moles of Fe2O3

= (3/1)*0.4897

= 1.469 mol

Molar mass of C = 12.01 g/mol

mass of C = number of mol * molar mass

= 1.469*12.01

= 17.64 g

Answer: 17.6 g

B)

Molar mass of C = 12.01 g/mol

mass of C = 38 g

mol of C = (mass)/(molar mass)

= 38/12.01

= 3.164 mol

According to balanced equation

mol of CO formed = moles of C

= 3.164 mol

Molar mass of CO,

MM = 1*MM(C) + 1*MM(O)

= 1*12.01 + 1*16.0

= 28.01 g/mol

mass of CO = number of mol * molar mass

= 3.164*28.01

= 88.62 g

Answer: 88.6 g

C)

Molar mass of Fe2O3,

MM = 2*MM(Fe) + 3*MM(O)

= 2*55.85 + 3*16.0

= 159.7 g/mol

mass of Fe2O3 = 6.4 g

mol of Fe2O3 = (mass)/(molar mass)

= 6.4/1.597*10^2

= 4.008*10^-2 mol

According to balanced equation

mol of Fe formed = (2/1)* moles of Fe2O3

= (2/1)*4.008*10^-2

= 8.015*10^-2 mol

Molar mass of Fe = 55.85 g/mol

mass of Fe = number of mol * molar mass

= 8.015*10^-2*55.85

= 4.476 g

Answer: 4.48 g