Question

A. How many moles of Al can be produced from 10.87 g of Ag?

Al(NO3)3(aq) + 3 Ag(s) → Al(s) + 3 AgNO3(aq)

B. What mass of O2 can be generated by the decomposition of 100.0 g of NaClO3?

2NaClO3(s) → 2 NaCl(s) + 3 O2(g)

C. What mass of Fe is generated when 100.0 g of Al are reacted?

Fe2O3(s) + 2 Al(s) → 2 Fe(s) + Al2O3(s)

Answer 1

A)

Molar mass of Ag = 107.9 g/mol

mass of Ag = 10.87 g
mol of Ag = (mass)/(molar mass)
= 10.87/1.079*10^2
= 0.1007 mol

According to balanced equation
mol of Al formed = (1/3)* moles of Ag
= (1/3)*0.1007
= 3.358*10^-2 mol
Answer: 3.358*10^-2 mol

B)

Molar mass of NaClO3,
MM = 1*MM(Na) + 1*MM(Cl) + 3*MM(O)
= 1*22.99 + 1*35.45 + 3*16.0
= 106.44 g/mol

mass of NaClO3 = 1*10^2 g
mol of NaClO3 = (mass)/(molar mass)
= 1*10^2/1.064*10^2
= 0.9395 mol

According to balanced equation
mol of O2 formed = (3/2)* moles of NaClO3
= (3/2)*0.9395
= 1.409 mol


Molar mass of O2 = 32 g/mol


mass of O2 = number of mol * molar mass
= 1.409*32
= 45.10 g
Answer: 45.10 g

C)

Molar mass of Al = 26.98 g/mol

mass of Al = 1*10^2 g
mol of Al = (mass)/(molar mass)
= 1*10^2/26.98
= 3.706 mol

According to balanced equation
mol of Fe formed = moles of Al
= 3.706 mol


Molar mass of Fe = 55.85 g/mol


mass of Fe = number of mol * molar mass
= 3.706*55.85
= 2.07*10^2 g
Answer: 2.070*10^2 g