Question

A metal has an atomic weight of 24 . When it reacts with a non-metal of atomic weight 80, it does so in a ratio of 1 atom to 2 atoms, respectively. With this information, how many grams of non-metal will combine with \(33.3\) g of metal. If \(1 \mathrm{~g}\) of metal is reacted with \(5 \mathrm{~g}\) of non-metal, find the amount of product produced.

Short answer

The molecular weight of the compound is 184. When 33.3 g of metal reacts with the non-metal, 111 g of non-metal is required. When 1 g of metal reacts with 5 g of non-metal, 7.67 g of product is produced.

Step-by-step solution

Determine Molecular Weight of Compound

First, we need to find the molecular weight of the compound they create. The metal reacts in a ratio of 1:2 with the non-metal, so we need to multiply the atomic weight of non-metal by 2 Molecular weight of compound = (Atomic weight of metal) + 2 * (Atomic weight of non-metal)

Calculate Molecular Weight

Now, we can plug in the given atomic weights of the metal and non-metal and calculate the molecular weight: Molecular weight = (24) + 2 * (80) Molecular weight = 24 + 160 Molecular weight = 184

Calculate Moles of Metal

To find out how much of the non-metal will react with 33.3 g of the metal, first we need to find the moles of the metal. Moles can be calculated using the formula: Moles = mass (grams) / atomic weight moles of metal = 33.3 g / 24

Calculate Mass of Non-metal

Since the metal reacts in a 1:2 ratio with the non-metal, we can multiply the moles of the metal by 2 to find the moles of the non-metal, and then calculate the mass of the non-metal: moles of non-metal = 2 * (moles of metal) mass of non-metal = moles of non-metal * atomic weight of non-metal

Determine Amount of Product Produced

Lastly, we are asked to find the amount of product produced when 1 g of metal reacts with 5 g of non-metal. We can assume that the reaction is complete. Since 1 g of metal is less than the amount required to react with 5 g of non-metal, the non-metal will be in excess, and the metal will be the limiting reagent. Therefore, the amount of product produced will be determined by the moles of the metal in the reaction: moles of metal (from 1 g) = 1 g / 24 mass of product produced = moles of metal (from 1 g) * molecular weight of product

Key concepts

Molecular Weight Calculation

Understanding the molecular weight calculation is foundational in chemistry, particularly when you're dealing with reactions and compounds. Molecular weight, also known as molecular mass, is the sum of the atomic weights of all the atoms in a molecule. It's expressed in atomic mass units (amu).

To calculate the molecular weight of a compound, you need to know the atomic weights of each element involved and the number of atoms of each element in the molecule. For example, if a compound is formed by one atom of metal (with an atomic weight of 24 amu) reacting with two atoms of a non-metal (with an atomic weight of 80 amu), the calculation would be as follows:

• Atomic weight of metal: 24 amu
• Atomic weight of non-metal: 80 amu x 2 atoms = 160 amu
• Molecular weight of the compound: 24 amu (metal) + 160 amu (non-metal) = 184 amu

This sum gives you the molecular weight of the newly formed compound. In a balanced chemical equation, this value helps you figure out how much of each reactant is needed and what amount of product is formed.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and the macroscopic world we live in. One mole equals Avogadro's number (\(6.022 \times 10^{23}\) entities), which is roughly the number of atoms in 12 grams of carbon-12. When we say we have one mole of a substance, it means we have \(6.022 \times 10^{23}\) of its fundamental units (atoms, molecules, ions, etc.).

To relate mass to moles, you use the formula:
\[\begin{equation}\text{Moles} = \frac{\text{mass (g)}}{\text{atomic or molecular weight (g/mol)}}\end{equation}\]For instance, if you have 33.3 grams of a metal with an atomic weight of 24 g/mol, you can calculate moles of metal as:
\[\begin{equation}\text{moles of metal} = \frac{33.3 \text{ g}}{24 \text{ g/mol}}\end{equation}\]This calculation tells you how many 'packets' of metal atoms you have, which is crucial when determining how the metal reacts with another element to form a compound. Understanding moles allows chemists to 'count' atoms by weighing them and to predict the mass of substances involved in reactions.

Limiting Reagent

In a chemical reaction, the limiting reagent is the reactant that gets used up first, thus determining the maximum amount of product that can be formed. Identifying the limiting reagent is crucial because it dictates the completion of the reaction; excess reagents are those with quantities greater than what's required to completely react with the limiting reagent.

To find the limiting reagent, you must perform mole-based comparisons using the balanced chemical equation. Considering the reaction between our metal and non-metal, if you have 1 gram of metal and 5 grams of non-metal, you would do the following to identify the limiting reagent:

• Calculate moles of metal: \[\begin{equation}\text{moles of metal} = \frac{1 \text{ g}}{24 \text{ g/mol}}\end{equation}\]
• Calculate moles of non-metal necessary to react with these moles of metal, based on the given reaction stoichiometry.
• Compare this with the actual moles of non-metal present (calculated from its mass and molar mass).

If the actual moles of non-metal are greater than what is required, then metal is the limiting reagent. Knowing this helps you predict the maximum amount of product that can be formed from given quantities of reactants, which is a fundamental aspect of stoichiometry.