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Assume that 8.5 L of iodine gas (I2) are produced at STP according to the following balanced equation:

2KI (aq) + Cl2 (g) --> 2KCl (aq) + I2 (g)

a ) How many moles of KI were used?

b ) How many grams of KI were used

any help is appreciated:(((

User TabeaKischka
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2 Answers

21 votes
21 votes

molar mass I equal to the lenses of gravity

User Sarme
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27 votes
27 votes

Answer:

a) 0.799 mol (3 s.f.)

b) 133 g (3 s.f.)

Step-by-step 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.)

Assume that 8.5 L of iodine gas (I2) are produced at STP according to the following-example-1
Assume that 8.5 L of iodine gas (I2) are produced at STP according to the following-example-2
User Cglacet
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