Question

Consider the reaction between solid iron(III) oxide and carbon monoxide gas to form carbon dioxide gas and solid iron (11,111) oxide. If carbon monoxide is present in excess, determine the amount of iron(III) oxide needed to produce 2.21x1024 iron (11,HI) oxide molecules. O 1.83 moles 3.67 moles 07.34 moles O 11.0 moles O 5.50 moles

Answer 1

sol:-

• 1 mole = 6.022 * 1023

1):- data provided in the question is :-

• molecule of Fe = 2.21 * 1024

moles of Fe = (molecules of Fe)/(molecules in 1mole)

= (2.21 * 1024)/(6.022 * 1023)

= 3.67 mole

reaction :-

Fe2O3 + CO $\rightarrow$ 2Fe + 3CO2

2 mole of Fe produced from 1 mole of Fe2O3

3.67 mole of Fe will produced from (1/2)*3.67 of Fe2O3

= 1.835 mole of Fe2O3

mole of Fe2O3 = 1.835 moles

2):-data provided in the question is :-

• mole of A = 0.50 mole
• mole of B = 0.60 mole
• mole of C = 0.90 mole

​​​​​​​reaction is :-

A + 2B + 3C $\rightarrow$ 2D + E

from the reaction :-

2 mole of B react with 1 mole of A so,

0.60 mole of B will react with (1/2)*0.60 mole of A

= 0.30 mole of A

then,

2 mole of B react with 3 mole C so,

0.60 mole of B will react with (3/2)*0.60 mole of C

= 0.90 mole of C

amount of A require = 0.30 mole

amount of B require = 0.60 mole

amount of C require = 0.90 mole

so,

A is in excess amount( 0.50 - 0.30 = 0.20) and other reactant is in less amount . so,

limiting reagent = B and C

3):-sol:-

• atomic mass of C = 12 g/mol.
• atomic mass of H = 1 g/mol.
• atomic mass of O = 16 g/mol.
• molecular mass of CH4 = (12 + 4*1) g/mol = 16 g/mol.
• molecular mass of C3H8 = (3*12 + 8*1)g/mol. = 44 g/mol.
• molecular mass of CO2 = (12 + 2*16) g/mol. = 44 g/mol.
• molecular mass of O2 = (2*16)g/mol = 32 g/mol

total mass of mixture of CH4 and C3H8 = 309 g

• amount of CH4 = 309(43.6/100) g = 309 * 0.436 g = 134.724 g
• amount of C3H8 = 309(56.4/100) g = 309 * 0.564 g = 174.276 g
• amount of O2​​​​​​​ is in excess.

​​​​​​​reaction :-

CH4 + 2O2 $\rightarrow$ CO2 + 2H2O -(1)

C3H8 + 5O2 $\rightarrow$ 3CO2 + 4H2O -(2)

from reaction (1):-

16 g of CH4 produce 44 g of CO2 then,

134.724 g of CH4 produce (44/16)*134.724 g of CO2

= 370.491 g of CO2

from reaction (2):-

44 g of C3H8 produce (3*44) g of CO2 then,

174.276 g of C3H8 produce (3*44/44)*174.276 g of CO2

= 522.828 g of CO2

then total amount of CO2 = (370.491 + 522.828) g = 893.319