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(a) What is the difference between adding a subscript 2 to the end of the formula for \(\mathrm{CO}\) to give \(\mathrm{CO}_{2}\) and adding a coefficient in front of the formula to give 2 CO? (b) Is the following chemical equation, as written, consistent with the law of conservation of mass? \(3 \mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{~s})+2 \mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l)\) Why or why not?

Short Answer

Expert verified
(a) Adding a subscript 2 to the end of the formula for \(\mathrm{CO}\) to give \(\mathrm{CO}_2\) changes the composition of the molecule, creating carbon dioxide. Adding a coefficient 2 in front of the formula to give \(2\mathrm{CO}\) means there are two molecules of carbon monoxide. (b) The given chemical equation is consistent with the law of conservation of mass, as the number of atoms of each element is equal on both sides of the equation.

Step by step solution

01

Part (a) - Difference between adding a subscript and adding a coefficient

In a chemical formula, a subscript indicates the number of atoms of a specific element in a molecule, while a coefficient specifies the number of molecules involved in the reaction. When a subscript is added to a chemical formula, such as adding a subscript 2 to the end of the formula for \(\mathrm{CO}\) to give \(\mathrm{CO}_2\), it changes the composition of the molecule. In this case, the molecule would now consist of one carbon atom and two oxygen atoms, forming carbon dioxide. On the other hand, adding a coefficient in front of a chemical formula, such as adding 2 in front of the formula \(\mathrm{CO}\) to give \(2\mathrm{CO}\), signifies that there are two molecules of carbon monoxide, each consisting of one carbon atom and one oxygen atom.
02

Part (b) - Checking if the chemical equation is consistent with the law of conservation of mass

We can check if the given chemical equation is consistent with the law of conservation of mass by counting the number of atoms of each element on both sides of the equation and making sure they are equal. The given chemical equation is: \[3 \mathrm{Mg(OH)_2(s)}+2 \mathrm{H_3PO_4(aq)} \longrightarrow \mathrm{Mg_3(PO_4)_2(s)}+6 \mathrm{H_2O(l)}\] Let's count the atoms of each element on both sides of the equation: Left-hand side of the equation: - Mg: \(3\) (from \(\mathrm{Mg(OH)_2}\)) - O: \(6\) (from \(\mathrm{Mg(OH)_2}\)) - H: \(12\) (from \(\mathrm{Mg(OH)_2}\)) + \(6\) (from \(\mathrm{H_3PO_4}\)) = \(18\) - P: \(2\) (from \(\mathrm{H_3PO_4}\)) Right-hand side of the equation: - Mg: \(3\) (from \(\mathrm{Mg_3(PO_4)_2}\)) - O: \(8\) (from \(\mathrm{Mg_3(PO_4)_2}\)) + \(6\) (from \(\mathrm{H_2O}\)) = \(14\) - H: \(12\) (from \(\mathrm{H_2O}\)) - P: \(2\) (from \(\mathrm{Mg_3(PO_4)_2}\)) Comparing the number of atoms of each element on both sides, we can see that they are equal. Therefore, the given chemical equation is consistent with the law of conservation of mass.

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Key Concepts

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

Subscripts in Chemistry
In chemistry, subscripts are used to indicate the number of atoms of a particular element within a molecule. They are written as small numbers to the right and slightly below the chemical symbols in a formula. For example, in the molecule \( \text{CO}_2 \), the subscript 2 suggests that there are two oxygen atoms bonded with one carbon atom to form carbon dioxide.
This small number is crucial because it tells us the actual composition of the molecule, confirming what substances are present and in what proportion.
When the subscript is altered, it changes the nature of the molecule completely. For instance, \( \text{CO} \) (carbon monoxide) contains one carbon and one oxygen, while \( \text{CO}_2 \) indicates a different compound—carbon dioxide—due to the subscript.
  • Subscripts are part of the molecular structure.
  • They help determine the molecule's identity and properties.
Understanding subscripts is essential because even a small change can indicate a vastly different substance.
Coefficients in Reactions
Coefficients in chemical reactions denote the number of molecules or moles involved in a reaction. They are written as regular-sized numbers in front of chemical formulas in an equation. For example, the equation \( 3 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O} \) shows coefficients 3, 1 (implied), and 2 respectively.
The coefficient informs us about the ratio in which reactants combine and products form. It's vital for balancing chemical equations and ensuring that the equation adheres to the law of conservation of mass.
  • Coefficients are not part of the molecular structure; they belong to the reaction as a whole.
  • They provide a means to balance the equation and reflect real-world conditions.
Adjusting coefficients alters the number of molecules present but keeps the molecular structure consistent.
Conservation of Mass
The law of conservation of mass is a fundamental principle in chemistry, stating that mass in a closed system must remain constant over time. Consequently, in any chemical reaction, the mass of the reactants must equal the mass of the products.
This principle is crucial for writing and balancing chemical equations, ensuring that atoms are neither lost nor gained but simply rearranged.
  • Mass is conserved in chemical reactions.
  • Balancing equations respects this law by equalizing atom counts on both sides.
Consider the example equation: \( 3 \mathrm{Mg(OH)_2} + 2 \mathrm{H_3PO_4} \rightarrow \mathrm{Mg_3(PO_4)_2} + 6 \mathrm{H_2O} \). Checking both sides, we see equal atoms of each element, demonstrating mass conservation.

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Most popular questions from this chapter

Hydrogen sulfide is an impurity in natural gas that must be removed. One common removal method is called the Claus process, which relies on the reaction: $$ 8 \mathrm{H}_{2} \mathrm{~S}(g)+4 \mathrm{O}_{2}(g) \longrightarrow \mathrm{S}_{8}(l)+8 \mathrm{H}_{2} \mathrm{O}(g) $$ Under optimal conditions the Claus process gives \(98 \%\) yield of \(\mathrm{S}_{8}\) from \(\mathrm{H}_{2} \mathrm{~S}\). If you started with \(30.0\) grams of \(\mathrm{H}_{2} \mathrm{~S}\) and \(50.0\) grams of \(\mathrm{O}_{2}\), how many grams of \(\mathrm{S}_{8}\) would be produced, assuming \(98 \%\) yield?

A particular coal contains \(2.5 \%\) sulfur by mass. When this coal is burned at a power plant, the sulfur is converted into sulfur dioxide gas, which is a pollutant. To reduce sulfur dioxide emissions, calcium oxide (lime) is used. The sulfur dioxide reacts with calcium oxide to form solid calcium sulfite. (a) Write the balanced chemical equation for the reaction. (b) If the coal is burned in a power plant that uses 2000 tons of coal per day, what mass of calcium oxide is required daily to eliminate the sulfur dioxide? (c) How many grams of calcium sulfite are produced daily by this power plant?

(a) Define the terms theoretical yield, actual yield, and percent yield. (b) Why is the actual yield in a reaction almost always less than the theoretical yield? (c) Can a reaction ever have \(110 \%\) actual yield?

Vanillin, the dominant flavoring in vanilla, contains \(\mathrm{C}\), \(\mathrm{H}\), and \(\mathrm{O} .\) When \(1.05 \mathrm{~g}\) of this substance is completely combusted, \(2.43 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.50 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are produced. What is the empirical formula of vanillin?

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(55.3 \% \mathrm{~K}, 14.6 \% \mathrm{P}\), and \(30.1 \% \mathrm{O}\) (b) \(24.5 \% \mathrm{Na}, 14.9 \% \mathrm{Si}\), and \(60.6 \% \mathrm{~F}\) (c) \(62.1 \% \mathrm{C}, 5.21 \% \mathrm{H}, 12.1 \% \mathrm{~N}\), and \(20.7 \% \mathrm{O}\)

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