F.4 CHEM

1. a)
2Cu + O2 → 2CuO
Molar ratio Cu : CuO = 2 : 2 = 1 : 1

Molar mass of Cu = 63.5 g/mol
Molar mass of CuO = 63.5 + 16 = 79.5

No. of moles of Cu reacted = 5/63.5 = 0.0787 mol
No. of moles of CuO formed = 0.0787 mol
Mass of CuO formed = 0.0787 * 79.5 = 6.26 g ...... (Answer)

b)
The copper contains some impurities which does not reacted.
Or: The copper is not completely reacted.


2.
Ca + 2H2O → Ca(OH)2 + H2
Mole ratio Ca : H2 = 1 : 1

Molar mass of Ca = 40 g/mol
Molar mass of H2 = 1*2 = 2 g/mol

No. of moles of Ca used = 24.6/40 = 0.615 mol
No. of moles of H2 formed = 0.615 mol
Mass of H2 formed = 0.615 * 2 = 1.23 g ...... (Answer)


3.
Molar mass of Ar = 40 g/mol
No. of moles of Ar = 64/40 = 1.6 mol
No. of Ar atoms = 1.6 * (6.02 * 10²³) = 9.63 * 10²³ ...... (Answer)

1a. 2Cu(s) + O2(g) --> 2CuO(s)
mole of Cu = 5/63.5
from the equation, 2 mole Cu produced 2 mole CuO
then 5/63.5 mole Cu produced 5/63.5 mole CuO
therefore, mass of oxide formed = 5/63.5 x (63.5+16) = 6.26g
1b. There may have loss of material during the experiment

2. Ca + H2O --> CaO + H2
mole of Ca used = 24.6/40.1
from the equation, 1 mole Ca form 1 mole H2
24.6/40.1 mole Ca form 24.6/40.1 mole H2
therefore, mass of H2 evolved = 24.6/40.1 x (1x2)
= 1.23g

3. no.of mole of 64g Ar = 64/40 = 1.6
By the Avogadro Constant, 1 mole contains 6.02 x 10^23 atom
therefore, no. of atoms that Ar contained = 1.6 x 6.02 x 10^23 = 9.632x10^23

1. 2Cu (s) + O2 (g) -->> 2CuO (s)
a. relative mass of Cu = 63.5
number of mole of Cu burnt = 5 / 63.5 = 0.07874mol
By equation, number of mole of CuO formed = 0.7874mol
relative mass of CuO = 63.5+18 = 81.5
mass of CuO = 81.5 x 0.07874 = 6.42 gram