Question #dc373

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Question #dc373

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Answer 1 · 2014-01-25T02:55:30

To get the experimental molar ratio, you divide the moles of each reactant that you actually used in the experiment by each other.

Explanation:

EXAMPLE 1

Consider the reaction: $2Al + {3I}_{2} \to {2AlI}_{3}$ $"2Al" + "3I"_2 → "2AlI"_3$

What is the experimental molar ratio of $Al$ $"Al"$ to $I_{2}$ $"I"_2$ if 1.20 g $Al$ $"Al"$ reacts with 2.40 g $I_{2}$ $"I"_2$ ?

Solution

Step 1: Convert all masses into moles.

$1.20 g Al \times \frac{1 mol Al}{26.98 g Al} = 0.044 48 mol Al$ $1.20 cancel("g Al") × "1 mol Al"/(26.98 cancel("g Al")) = "0.044 48 mol Al"$

$2.40 g I₂ \times \frac{{1 mol I}_{2}}{253.8 g I₂} = {0.009 456 mol I}_{2}$ $2.40 cancel("g I₂") × ("1 mol I"_2)/(253.8 cancel("g I₂")) = "0.009 456 mol I"_2$

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 $Al$ $"Al"$ to $I_{2}$ $"I"_2$ of $\frac{0.04448}{0.009456}$ $0.04448/0.009456$

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 $Al$ $"Al"$ to $I_{2}$ $"I"_2$ is then $\frac{0.04448}{0.009456} = \frac{4.70}{1}$ $0.04448/0.009456 = 4.70/1$ (3 significant figures)

The experimental molar ratio of $I_{2}$ $"I"_2$ to $Al$ $"Al"$ is $\frac{1}{4.70}$ $1/4.70$

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

${BaCl}_{2} (aq) + {2AgNO}_{3} (aq) \to 2AgCl(s) + {Ba(NO}_{3})_{2} (aq)$ $"BaCl"_2("aq") + "2AgNO"_3("aq") → "2AgCl(s)" + "Ba(NO"_3)_2("aq")$

She isolated 14.5 g of silver chloride. What was her experimental molar ratio of $AgCl$ $"AgCl"$ to ${BaCl}_{2}$ $"BaCl"_2$ ?

Solution

Step 1: Convert all masses into moles

$10.2 g BaCl₂ \times \frac{{1 mol BaCl}_{2}}{208.2 g BaCl₂} = {0.048 99 mol BaCl}_{2}$ $10.2 cancel("g BaCl₂") × ("1 mol BaCl"_2)/(208.2 cancel("g BaCl₂")) = "0.048 99 mol BaCl"_2$

$14.5 g AgCl \times \frac{1 mol AgCl}{143.3 g AgCl} = 0.1012 mol AgCl$ $14.5 cancel("g AgCl") × "1 mol AgCl"/(143.3 cancel("g AgCl")) = "0.1012 mol AgCl"$

Step 2: Calculate the molar ratios

The experimental molar ratio of $AgCl$ $"AgCl"$ to ${BaCl}_{2}$ $"BaCl"_2$ is $\frac{0.1012}{0.04899} = \frac{2.07}{1}$ $0.1012/0.04899 = 2.07/1$

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