d)
Mixed calculations
Example 6.4: Calculate the mass of hydrochloric acid
required for neutralizing 1 kg of NaOH
Solution: The neutralization reaction involved between
hydrochloric acid and sodium hydroxide is as follows :
| HCl(aq) |
+ |
NaOH(aq) |
—> |
NaCl(aq) |
+ |
H2O(l) |
| 1
mole 1+35.5 = 36.5 g |
|
1
mole 23+16+1 = 40 g |
|
|
|
|
Thus,
for neutralizing 40 g of NaOH the mass of HCl required is
36.5 g.
For neutralizing 1 kg or 1000 g of NaOH the mass of HCl
required is
36.5*1000/40 = 912.5 g
Example
6.5: In the reaction
2Na
( s ) + NaOH ( aq ) —> NaCl ( aq ) + H2O
calculate
the following:
(i) The maximum number of moles of sodium that can react
with 4 moles of water.
(ii) The mass of sodium hydroxide that would be produced
when 4.6 g of sodium reacts with excess of water.
(iii) The mass and volume at STP of hydrogen gas that would
be produced when 1.8 g of water reacts completely with sodium
metal.
Solution:
| 2Na(s) |
+ |
2H2O(l) |
—> |
2NaOH(aq) |
+ |
H2(g) |
| 2 moles 2*23=46g |
|
2 moles 2*18=36g |
|
2 moles 2*40=80g |
|
1 mole 1*2=2g |
(i)
From the equation it can be seen that
2 moles of water react with 2 moles of sodium
4 moles of water can react with a maximum of 4 moles of
sodium.
(ii) 46 g sodium reacts to produce 80 g sodium hydroxide
4.6 g sodium would produce 80*4.6/46= 8.0 g sodium hydroxide.
(iii) 6 g of water produces 2 g or 22.7 L of hydrogen at
STP
1.8 g of water would produce 2*1.8/36= 0.1 g of hydrogen
and 22.7*1.8/36= 1.135 L
of hydrogen at STP.
