Answer:
a) 0.799 mol (3 s.f.)
b) 133 g (3 s.f.)
Explanation:
2KI (aq) + Cl₂ (g) → 2KCl (aq) + I₂ (g)
1) Check if the equation has been balanced.
• The equation is already balanced in this question as the number of atoms of each element is equal on both sides.
Part (a)
2) Convert 8.5 L of Iâ‚‚ into number of moles.
At STP (standard temperature and pressure), 1 mole of gas occupies 22.4 L.
Amount of Iâ‚‚ gas produced
= 8.5 ÷22.4
= 0.37946 mol (5 s.f.)
3) Identify the relationship between Iâ‚‚ (g) produces and KI (aq) used in terms of mole ratio.
From the balanced equation, the mole ratio of KI (aq) used to the amount of Iâ‚‚ (g) produced is given as:
KI (aq): Iâ‚‚ (g)
= 2 :1
This is obtained by comparing the coefficients of KI (aq) and Iâ‚‚ (g) after the equation is balanced.
The mole ratio tells us that for every 1 mole of Iâ‚‚ (g) produced, 2 moles of KI (aq) is needed.
4) Calculate the amount of KI (aq) used.
1 mole of Iâ‚‚ (g) ----- 2 moles of KI (aq)
0.37946 mol of Iâ‚‚ (g) ----- 2(0.37946)= 0.79852 mol of KI (aq) (5 s.f.)
Thus, 0.799 mol of KI were used. (3 s.f.)
Part (b)
5) Convert number of moles into weight.
Mole ×Mr= Weight in grams
Mr stands for relative formula mass and can be calculated with the use of a periodic table.
Weight of KI (aq) used
= 0.79852 ×(39.1 +127)
= 0.79852 ×166.1
= 133 g (3 s.f.)
