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 I₂ into number of moles.
At STP (standard temperature and pressure), 1 mole of gas occupies 22.4 L.
Amount of I₂ gas produced
= 8.5 ÷22.4
= 0.37946 mol (5 s.f.)
3) Identify the relationship between I₂ (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 I₂ (g) produced is given as:
KI (aq): I₂ (g)
= 2 :1
This is obtained by comparing the coefficients of KI (aq) and I₂ (g) after the equation is balanced.
The mole ratio tells us that for every 1 mole of I₂ (g) produced, 2 moles of KI (aq) is needed.
4) Calculate the amount of KI (aq) used.
1 mole of I₂ (g) ----- 2 moles of KI (aq)
0.37946 mol of I₂ (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.)
