Question #dc373

EXAMPLE 1

Consider the reaction: #"2Al" + "3I"_2 → "2AlI"_3#

What is the experimental molar ratio of #"Al"# to #"I"_2# if 1.20 g #"Al"# reacts with 2.40 g #"I"_2#?

Solution

Step 1: Convert all masses into moles.

#1.20 cancel("g Al") × "1 mol Al"/(26.98 cancel("g Al")) = "0.044 48 mol Al"#

#2.40 cancel("g I₂") × ("1 mol I"_2)/(253.8 cancel("g I₂")) = "0.009 456 mol I"_2#

Step 2: Calculate the molar ratios

To calculate the molar ratios, you put the moles of one reactant over the moles of the other reactant.

This gives you a molar ratio of #"Al"# to #"I"_2# of #0.04448/0.009456#

Usually, you divide each number in the fraction by the smaller number of moles. This gives a ratio in which no number is less than 1.

The experimental molar ratio of #"Al"# to #"I"_2# is then #0.04448/0.009456 = 4.70/1# (3 significant figures)

The experimental molar ratio of #"I"_2# to #"Al"# is #1/4.70#

Note: It is not incorrect to divide by the larger number and express the above ratios as 1:0.213 and 0.213:1, respectively. It is just a matter of preference.

EXAMPLE 2

A student reacted 10.2 g of barium chloride with excess silver nitrate, according to the equation

#"BaCl"_2("aq") + "2AgNO"_3("aq") → "2AgCl(s)" + "Ba(NO"_3)_2("aq")#

She isolated 14.5 g of silver chloride. What was her experimental molar ratio of #"AgCl"# to #"BaCl"_2#?

Solution

Step 1: Convert all masses into moles

#10.2 cancel("g BaCl₂") × ("1 mol BaCl"_2)/(208.2 cancel("g BaCl₂")) = "0.048 99 mol BaCl"_2#

#14.5 cancel("g AgCl") × "1 mol AgCl"/(143.3 cancel("g AgCl")) = "0.1012 mol AgCl"#

Step 2: Calculate the molar ratios

The experimental molar ratio of #"AgCl"# to #"BaCl"_2# is #0.1012/0.04899 = 2.07/1#

Here is a video example:

video from: Noel Pauller