Answer to Question #155032 in General Chemistry for Dylan

Question

What is the mass of I needed to react with 5.35 grams of Al?

Answer 1

Q155032

What is the mass of I needed to react with 5.35 grams of Al?

Solution:

We will first write the reaction of Aluminium and iodine.

Aluminum exists as Al(s) in its elemental form. Iodine exists as I₂ (s) in its elemental form.

Aluminum is a metal and Iodine is a non-metal.

metal and non-metal combines to form ionic compounds.

Al belongs to III A group of the periodic table and Iodine belongs to VII A groups of the periodic table.

Al can give 3 electrons, so after giving 3 electrons to 3 atoms of I, it forms Al⁺³ ion

Iodine needs 1 electron for completing its outermost shell. It gains 1 electron and form

I^-1 ion.

Al ⁺³ and I ^-1 forms AlI₃

Al (s) + I₂ (s) ----> AlI₃ (s)

After balancing the reaction we have

2 Al (s) + 3 I₂ (s) ----> 2 AlI₃ (s)

Step 1: Convert 5.35 grams of Al to moles using the atomic mass of Al.

Atomic mass of Al = 26.982g/mol

$mol \space of \space Al = 5.35 \space g \space Al * \frac{1 \space mol Al}{26.982 \space g \space Al} = 0.1983 \space mol \space Al$ ;

Step 2: Using mole to mole ratio of Al and I₂ from the reaction find the moles of I₂.

The mole to mole ratio of Al and I₂ in the given reaction = 3 mol I₂: 2 mol Al.

$mol \space of Al = 0.1983 \space mol\space Al * \frac{3 mol \space I_2}{2\space mol Al } = 0.2974\space mol\space I_2$

Step 3: Using molar mass of I₂ convert 0.2974 mol I₂ to grams.

Atomic mass of I = 126.9g/mol

molar mass of I₂ = 2 * atomic mass of I = 2 * 126.9g/mol = 253.9g/mol

$grams \space of I_2 = 0.2974 \space mol \space I_2*\frac{253.8 \space g \space I_2}{1 \space mol \space I_2} = 75.48\space g \space of\space I_2$ ;

In question, we are given 5.35grams in 3 significant figures, so our final answer must also be in 3 significant figures.

Hence the mass of iodine needed to react with 5.35 grams of Al is 75.5grams.