# How to Solve Stoichiometry Problems Involving Mass and Balanced Equations?

• GLprincess02
In summary: It is a dimensional analysis setup. So by taking the mole fraction and multiplying it by the number of moles of aluminum nitrate, you can get the soln.
GLprincess02

## Homework Statement

Given the balanced equation 2Al(NO3)3 + 3Mg ==> 3Mg(NO3)2 + 2Al , if 54.1 g of aluminum nitrate is reacted, what mass of magnesium nitrate is produced?

2. The attempt at a solution
I've been given an entire packet dedicated to stoichiometry, but nothing in here covers this problem. I am unsure of how to even begin solving it. Maybe if I had help with the first few steps I could solve the rest.

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a good way to go about this problem is to convert the mass of aluminum nitrate into moles of aluminum nitrate, then you can figure out how many moles of magnesium (do you mean aluminum or magnesium nitrate) were produced.

Ok so I converted the grams of aluminum nitrate into moles and got approx. 0.2539 moles. I'm still confused about what to do next, though.

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GLprincess02 said:
Ok so I converted the grams of aluminum nitrate into moles and got approx. 0.2539 moles. I'm still confused about what to do next, though.

Now you need to use the mole fraction, which you get from the balanced reaction.

What is the mole fraction? Like, 2 moles of aluminum nitrate to 3 moles of magnesium nitrate?

GLprincess02 said:
What is the mole fraction? Like, 2 moles of aluminum nitrate to 3 moles of magnesium nitrate?

Yes.
(2mol Al nitrate/3mol Mg nitrate)

Ok, so how do I use this ratio in relation to the 0.2539 moles of aluminum nitrate? I guess I still don't really get all of the steps.

GLprincess02 said:
Ok, so how do I use this ratio in relation to the 0.2539 moles of aluminum nitrate? I guess I still don't really get all of the steps.

It is a dimensional analysis setup

mol Al nitrate(mole fraction)(convert mol Mg nitrate to g)=soln

So I take 0.2539 and multiply it by 2/3 first, right?

Never mind, I got the right answer. Thanks for the help. :)

## 1. What is stoichiometry and why is it important?

Stoichiometry is the study of the quantitative relationships between substances in a chemical reaction. It helps us determine the amount of reactants needed to produce a certain amount of products, and vice versa. This is important because it allows us to understand and predict the outcome of chemical reactions and optimize them for desired results.

## 2. How do I balance a chemical equation for stoichiometry?

To balance a chemical equation, you need to make sure that the number of atoms of each element is the same on both sides of the equation. This can be done by adjusting the coefficients (numbers in front of each compound) until the number of atoms on both sides are equal. It is important to note that you cannot change the subscripts in a chemical formula.

## 3. What is the mole concept in stoichiometry?

The mole concept in stoichiometry refers to the unit of measurement used to count the number of particles (atoms, molecules, etc.) in a substance. 1 mole of a substance is equal to its atomic/molecular mass in grams. This allows us to convert between mass and number of particles in a chemical reaction.

## 4. How do I use stoichiometry to calculate the limiting reactant?

The limiting reactant is the substance that is completely used up in a chemical reaction, thus limiting the amount of product that can be formed. To calculate the limiting reactant, you need to compare the amount of each reactant present to the balanced chemical equation. The reactant with the smallest amount (in moles) is the limiting reactant.

## 5. Can you provide an example of a stoichiometry problem and its solution?

Sure! Let's say we have the following reaction:
2H2 + O2 → 2H2O
If we have 4 moles of H2 and 6 moles of O2, how many moles of water will be produced?

To solve this, we first need to determine the limiting reactant.
For H2: 4 moles × 1 mol H2 ÷ 2 mol H2 = 2 moles H2
For O2: 6 moles × 1 mol O2 ÷ 1 mol O2 = 6 moles O2
Since we need 2 moles of H2 to react with 1 mole of O2, O2 is the limiting reactant.
Now, we can use the mole ratio to calculate the moles of water produced:
6 moles O2 × 2 mol H2O ÷ 1 mol O2 = 12 moles H2O
Therefore, 12 moles of water will be produced in this reaction.

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