# Basic Stoichiometry Post-lab Homework Exercises Rvsd 2 2011 Answers

Basic Stoichiometry Post-Lab Homework Exercises

## Basic Stoichiometry Post-Lab Homework Exercises

Stoichiometry is the study of the quantitative relationships between the reactants and products in a chemical reaction. In this article, we will review some basic concepts of stoichiometry and answer some post-lab homework exercises based on the PhET simulation "Reactants, Products, and Leftovers".

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## Basic Concepts of Stoichiometry

A chemical reaction is a process in which atoms or molecules (called reactants) are rearranged to form new substances (called products).

A chemical equation is a symbolic representation of a chemical reaction, showing the reactants, products, and their physical states.

A balanced chemical equation obeys the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. This means that the number and type of atoms on both sides of the equation must be equal.

A mole is a unit of measurement that represents 6.02 x 10 particles (atoms, molecules, ions, etc.) of a substance. The molar mass of a substance is the mass in grams of one mole of that substance.

The mole ratio is the ratio of the coefficients of two substances in a balanced chemical equation. It can be used to convert between the amounts (in moles) of different substances involved in a reaction.

The limiting reactant is the reactant that is completely consumed in a chemical reaction and determines the maximum amount of product that can be formed.

The excess reactant is the reactant that remains after the limiting reactant is used up in a chemical reaction. It does not affect the amount of product formed.

A synthesis reaction is a type of chemical reaction in which two or more simple substances combine to form a more complex substance.

A combustion reaction is a type of chemical reaction in which a substance reacts with oxygen to produce heat and light. The products are usually carbon dioxide and water.

A diatomic molecule is a molecule that consists of two atoms of the same element, such as hydrogen (H2), oxygen (O2), or nitrogen (N2).

A hydrocarbon is an organic compound that contains only carbon and hydrogen atoms, such as methane (CH4), ethane (C2H6), or propane (C3H8).

## Post-Lab Homework Exercises

The following exercises are based on the PhET simulation "Reactants, Products, and Leftovers", which can be accessed online at [PhET] or by googling "phet". The simulation allows you to explore different types of chemical reactions and see how the amounts of reactants and products change as the reaction proceeds. You can also play a game mode where you have to adjust the amounts of reactants to match a given amount of product.

To answer these exercises, you will need to use the concepts and formulas explained above. You will also need to balance some chemical equations, which can be done by following these steps:

Write down the unbalanced equation with the correct chemical formulas for each substance.

Count the number of atoms of each element on both sides of the equation.

Add coefficients (numbers) in front of each substance to make the number of atoms equal on both sides. Start with the most complex substance and work your way to the simplest ones. Do not change the subscripts (numbers) within each formula.

Check your work by counting the atoms again and making sure they are balanced.

Simplify your equation by dividing all coefficients by their greatest common factor, if possible.

Exercise 1:

For the reaction $$\textP_4 + \textCl_2 \rightarrow \textPCl_3$$ determine the correct lowest mole ratio. Solution:

To find the mole ratio, we need to balance the equation first. We can do this by following the steps above: 1. The unbalanced equation is: $$\textP_4 + \textCl_2 \rightarrow \textPCl_3$$ 2. The number of atoms of each element on both sides is: Element Reactants Products ------------------------------ P 4 1 Cl 2 3 3. To balance the equation, we can add coefficients as follows: $$\textP_4 + 6\textCl_2 \rightarrow 4\textPCl_3$$ 4. The number of atoms of each element on both sides is now: Element Reactants Products ------------------------------ P 4 4 Cl 12 12 The equation is balanced. 5. The equation cannot be simplified further, so this is the final balanced equation. The mole ratio is the ratio of the coefficients of any two substances in the equation. For example, the mole ratio of P4 to Cl2 is 1:6, the mole ratio of P4 to PCl3 is 1:4, and the mole ratio of Cl2 to PCl3 is 6:4 or 3:2. The lowest mole ratio is the one that has the smallest whole numbers, which in this case is 3:2 for Cl2 to PCl3. Exercise 2:

For the reaction $$\textC_3\textH_8 + \textO_2 \rightarrow \textCO_2 + \textH_2\textO$$ determine the mass of water produced when 14.0 g of propane (C3H8) is burned in excess oxygen. Solution:

To solve this problem, we need to use the concept of stoichiometry and the following formula: $$\textmass of product = \textmass of reactant \times \frac\textmole ratio of product to reactant\textmolar mass of reactant \times \textmolar mass of product$$ We also need to balance the equation first. We can do this by following the steps above: 1. The unbalanced equation is: $$\textC_3\textH_8 + \textO_2 \rightarrow \textCO_2 + \textH_2\textO$$ 2. The number of atoms of each element on both sides is: Element Reactants Products ------------------------------ C 3 1 H 8 2 O 2 3 3. To balance the equation, we can add coefficients as follows: $$\textC_3\textH_8 + 5\textO_2 \rightarrow 3\textCO_2 + 4\textH_2\textO$$ 4. The number of atoms of each element on both sides is now: Element Reactants Products ------------------------------ C 3 3 H 8 8 O 10 10 The equation is balanced. 5. The equation cannot be simplified further, so this is the final balanced equation. Now we can use the formula to find the mass of water produced. We have: - mass of propane (reactant) = 14.0 g - mole ratio of water (product) to propane (reactant) = 4:1 - molar mass of propane (reactant) = 44.1 g/mol - molar mass of water (product) = 18.0 g/mol Plugging these values into the formula, we get: $$\textmass of water = 14.0 \times \frac444.1 \times 18.0$$ $$\textmass of water = 11.4 \text g$$ Therefore, the mass of water produced when 14.0 g of propane is burned in excess oxygen is 11.4 g. Exercise 3:

For the reaction $$\textN_2 + \textH_2 \rightarrow \textNH_3$$ determine the amount of ammonia (NH3) produced when 28.0 g of nitrogen (N2) reacts with 25.0 g of hydrogen (H2). Solution:

To solve this problem, we need to use the concept of stoichiometry and the following formula: $$\textamount of product (in moles) = \textamount of reactant (in moles) \times \textmole ratio of product to reactant$$ We also need to balance the equation first. We can do this by following the steps above: 1. The unbalanced equation is: $$\textN_2 + \textH_2 \rightarrow \textNH_3$$ 2. The number of atoms of each element on both sides is: Element Reactants Products ------------------------------ N 2 1 H 2 3 3. To balance the equation, we can add coefficients as follows: $$\textN_2 + 3\textH_2 \rightarrow 2\textNH_3$$ 4. The number of atoms of each element on both sides is now: Element Reactants Products ------------------------------ N 2 2 H 6 6 The equation is balanced. 5. The equation cannot be simplified further, so this is the final balanced equation. Now we can use the formula to find the amount of ammonia produced. We have: - amount of nitrogen (reactant) = 28.0 g - amount of hydrogen (reactant) = 25.0 g - mole ratio of ammonia (product) to nitrogen (reactant) = 2:1 - molar mass of nitrogen (reactant) = 28.0 g/mol - molar mass of hydrogen (reactant) = 2.0 g/mol To use the formula, we need to convert the masses of the reactants to moles first. We can do this by dividing the mass by the molar mass: - amount of nitrogen (in moles) = 28.0 / 28.0 = 1.0 mol - amount of hydrogen (in moles) = 25.0 / 2.0 = 12.5 mol Plugging these values into the formula, we get: $$\textamount of ammonia = 1.0 \times \frac21$$ $$\textamount of ammonia = 2.0 \text mol$$ However, this is not the final answer, because we have to consider the limiting reactant. The limiting reactant is the reactant that is completely consumed in a chemical reaction and determines the maximum amount of product that can be formed. To find the limiting reactant, we can compare the amounts of reactants in terms of their mole ratios in the balanced equation. For example, according to the balanced equation, for every mole of nitrogen, we need three moles of hydrogen to react completely. However, we have more than three times as much hydrogen as