How many moles of Al(NO3)3 are produced when 15.87 grams of HNO3 are reacted with excess Al(OH)3?

Question

How many moles of Al(NO₃)₃ are produced when 15.87 grams of HNO₃ are reacted with excess Al(OH)₃? Round your answer to three digits after the decimal point.

* Al(OH)₃ + 3 HNO₃ = Al(NO₃)₃ + 3 H₂O

Answer 1

To determine the number of moles of Al(NO₃)₃ produced when 15.87 grams of HNO₃ react with excess Al(OH)₃, we need to use the balanced chemical equation and perform a mole-to-mole conversion.

The balanced chemical equation is:

Al(OH)₃ + 3HNO₃ → Al(NO₃)₃ + 3 H₂O

According to the equation, the stoichiometric ratio between HNO3 and Al(NO₃)₃ is 3:1. This means that for every 3 moles of HNO₃ reacted, 1 mole of Al(NO₃)₃ is produced.

First, let's calculate the number of moles of HNO₃:

Molar mass of HNO₃ = 1 * 1.01 (H) + 1 * 14.01 (N) + 3 * 16.00 (O)

= 63.02 g/mol

Number of moles of HNO₃ = Mass of HNO₃ / Molar mass of HNO₃

= 15.87 g / 63.02 g/mol

≈ 0.2518 moles

Since the stoichiometric ratio between HNO₃ and Al(NO₃)₃ is 3:1, the number of moles of Al(NO3)3 produced will be one-third of the number of moles of HNO₃:

Number of moles of Al(NO₃)₃ = (1/3) * Number of moles of HNO₃

= (1/3) * 0.2518 moles

≈ 0.0839 moles

Therefore, approximately 0.084 moles of Al(NO₃)₃ are produced when 15.87 grams of HNO₃ are reacted with excess Al(OH)₃.

Answer 2

Hi Sydney,

This is a great question dealing with stoichiometry.

When I solve a stoichiometry question, I always like to start by answering three simple questions:

1. What measurement is given?
2. What measurement are we trying to find?
3. What conversions do we need to use to get to our answer?

With the problem you've provided, the answers to the questions would be the following:

1. We are given 15.87 g of HNO₃
2. We are looking for ____ moles of Al(NO₃)₃
3. Conversions we will need are the molar mass of HNO₃ (found from the periodic table= 63.01g/mole) and the mole ratios from HNO₃ to Al(NO₃)₃ (found from the balanced equation= 3 moles HNO₃/ 1 mole Al(NO₃)₃).

Hurray! We now have all the information we need to solve the equation! We just plug it in as follows:

15.87 g HNO₃ x 1 mole HNO₃/ 63.01g x 1 mole Al(NO₃)₃ / 3 moles HNO₃ = 0.084 moles Al(NO₃)_3.