For each of the following balanced chemical equations, write all possible mole ratios: a. \(2 \mathrm{Ca}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{CaO}\) b. \(\mathrm{Mg}+2 \mathrm{HF} \longrightarrow \mathrm{MgF}_{2}+\mathrm{H}_{2}\)

Short Answer

Expert verified
a) 3 mole ratios: \[ \text{Ca:O}_2=2:1, \text{Ca:CaO}=1:1, \text{O}_2:\text{CaO}=1:2.\] b) 6 mole ratios: \[ \text{Mg:HF}=1:2, \text{Mg:MgF}_2=1:1, \text{Mg:H}_2=1:1, \text{HF:MgF}_2=2:1, \text{HF:H}_2=2:1, \text{MgF}_2:\text{H}_2=1:1.\]

Step by step solution

01

Understanding Mole Ratios

Mole ratios are the ratios between the amounts in moles of any two compounds involved in a chemical reaction. The coefficients in the balanced chemical equation represent these ratios.
02

Write Mole Ratios for Equation a

For the balanced equation: \(2 \text{Ca} + \text{O}_2 \rightarrow 2 \text{CaO}\), the mole ratios can be written by comparing the coefficients: 1. \[\text{Ratio of Ca to O}_2: \frac{2 \text{moles of Ca}}{1 \text{mole of O}_2}\]2. \[ \text{Ratio of Ca to CaO: \frac{2 \text{moles of Ca}}{2 \text{moles of CaO}}=1}\]3. \[ \text{Ratio of O}_2 \text{ to CaO: \frac{1 \text{mole of O}_2}{2 \text{moles of CaO}}}\]
03

Write Mole Ratios for Equation b

For the balanced equation: \( \text{Mg} + 2 \text{HF} \rightarrow \text{MgF}_2 + \text{H}_2 \), the mole ratios can be written by comparing the coefficients: 1. \[\text{Ratio of Mg to HF: \frac{1 \text{mole of Mg}}{2 \text{moles of HF}}\]2. \[ \text{Ratio of Mg to MgF}_2: \frac{1 \text{mole of Mg}}{1 \text{mole of MgF}_2}\]3. \[ \text{Ratio of Mg to H}_2: \frac{1 \text{mole of Mg}}{1 \text{mole of H}_2}\]4. \[ \text{Ratio of HF to MgF}_2: \frac{2 \text{moles of HF}}{1 \text{mole of MgF}_2}\]5. \[ \text{Ratio of HF to H}_2: \frac{2 \text{moles of HF}}{1 \text{mole of H}_2}\]6. \[ \text{Ratio of MgF}_2 to H}_2: \frac{1 \text{mole of MgF}_2}{1 \text{mole of H}_2}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balanced Chemical Equations
Chemical reactions are represented by chemical equations. These equations show the reactants (the starting substances) and the products (the substances formed in the reaction). To reflect the conservation of mass, the number of atoms for each element must be the same on both sides of the equation. This is called a balanced chemical equation.
For example, let's look at the equation: \[2\text{Ca} + \text{O}_2 \longrightarrow 2\text{CaO}\] This means that 2 moles of calcium (Ca) react with 1 mole of oxygen (\text{O}_2) to produce 2 moles of calcium oxide (CaO). Each side of the balanced equation has 2 calcium atoms and 2 oxygen atoms.
Writing balanced equations ensures that we adhere to the law of conservation of mass, which states that mass cannot be created or destroyed in a chemical reaction. Balanced equations are essential for accurate stoichiometric calculations, which is the next key concept we'll explore.
Stoichiometry
Stoichiometry is the area of chemistry that deals with the relative quantities of reactants and products in chemical reactions. It uses the balanced chemical equations to determine the relationships between the amounts in moles of reactants and products.
The coefficients in a balanced equation allow chemists to calculate how much of each substance is needed or produced. These coefficients represent mole ratios, which are crucial for solving stoichiometric problems.
For instance, in the equation \[\text{Mg} + 2\text{HF} \longrightarrow \text{MgF}_2 + \text{H}_2\] \text{Mg} and \text{HF} react in a 1:2 ratio. So, if you start with 1 mole of \text{Mg}, you'll need 2 moles of \text{HF} to completely react with it. This understanding is essential to figure out the quantities involved in chemical reactions.
Moles
A mole is a fundamental concept in chemistry that denotes a quantity of particles, typically atoms or molecules. One mole contains exactly \(6.022\times10^{23}\text{particles}\) (Avogadro's number).
The use of moles allows chemists to count atoms and molecules by weighing them. This is because chemical reactions, as represented in balanced equations, occur in definite mole ratios. By knowing the molar mass (mass of one mole) of each substance, you can convert between mass and moles. For example, if you have 12 grams of carbon (\text{C}), and the molar mass of carbon is 12 g/mol, then you have 1 mole of carbon atoms.
Understanding moles and mole ratios is crucial for interpreting and predicting the outcomes of chemical reactions, which ultimately lets us calculate the quantities of reactants needed and products formed in any given reaction.

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Most popular questions from this chapter

Magnesium is obtained from sea water. \(\mathrm{Ca}(\mathrm{OH})_{2}\) is added to sea water to precipitate \(\mathrm{Mg}(\mathrm{OH})_{2} .\) The precipitate is filtered and reacted with HCl to produce \(\mathrm{MgCl}_{2} .\) The \(\mathrm{MgCl}_{2}\) is electrolyzed to produce Mg and \(\mathrm{Cl}_{2} .\) If 185.0 \(\mathrm{g}\) of magnesium are recovered from \(1000 . \mathrm{gMgCl}_{2},\) what is the percentage yield for this reaction?

Relating Ideas The chemical equation is a good source of information concerning a reaction. Explain the relationship between the actual yield of a reaction product and the chemical equation of the product.

Sodium chloride is produced from its elements through a synthesis reaction. What mass of each reactant would be required to produce 25.0 \(\mathrm{mol}\) of sodium chloride?

Calculate the indicated quantity for each of the various chemical reactions given: a. theoretical yield \(=20.0 \mathrm{g},\) actual yield \(=15.0 \mathrm{g}\) percentage yield \(=?\) b. theoretical yield \(=1.0 \mathrm{g},\) percentage yield \(=90.0 \%\) actual yield \(=?\) c. theoretical yield \(=5.00 \mathrm{g},\) actual yield \(=4.75 \mathrm{g}\) percentage yield \(=?\) d. theoretical yield \(=3.45 \mathrm{g}\) , percentage yield = 48.0\(\%\) , actual yield =?

During lightning flashes, nitrogen combines with oxygen in the atmosphere to form nitrogen monoxide, NO, which then reacts further with \(\mathrm{O}_{2}\) to produce nitrogen dioxide, NO. a. What mass of \(\mathrm{NO}_{2}\) , is formed when NO reacts with 384 \(\mathrm{g} \mathrm{O}_{2} ?\) b. How many grams of NO are required to react with this amount of \(\mathrm{O}_{2} ?\)

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