Question

For each of the following unbalanced equations, calculate the mass of each product that could be produced by complete reaction of \(1.55 \mathrm{g}\) of the reactant indicated in boldface. a. \(\mathbf{C S}_{2}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{SO}_{2}(g)\) b. \(\mathbf{N a N O}_{3}(s) \rightarrow \operatorname{NaNO}_{2}(s)+\mathrm{O}_{2}(g)\) c. \(\mathrm{H}_{2}(g)+\mathbf{M} \mathbf{n} \mathbf{O}_{2}(s) \rightarrow \mathrm{MnO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathbf{B} \mathbf{r}_{2}(l)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{BrCl}(g)\)

Short answer

a. Mass of CO₂ = 0.8972 g and Mass of SO₂ = 2.6099 g b. Mass of NaNO₂ = 1.2586 g and Mass of O₂ = 0.2918 g c. Mass of MnO = 1.2652 g and Mass of H₂O = 0.6412 g d. Mass of BrCl = 2.2379 g

Step-by-step solution

Balance the chemical equation

\(CS_2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g)\)

Convert mass to moles for \(CS_2\)

Given mass of \(CS_2 = 1.55 g\). The molar mass of \(CS_2 = 12 + 2 \times 32 = 76 g/mol\). Therefore, moles of \(CS_2 = \frac{1.55}{76} = 0.02039 mol\)

Calculate moles of products

Using the balanced equation: \(1mol\; CS_2 \rightarrow 1mol\; CO_2\) and \(1mol\; CS_2 \rightarrow 2mol\; SO_2\) Moles of \(CO_2 = 0.02039 mol\) Moles of \(SO_2 = 2 \times 0.02039 mol = 0.04078 mol\)

Convert moles of products to grams

The molar mass of \(CO_2 = 12 + 2 \times 16 = 44 g/mol\) The molar mass of \(SO_2 = 32 + 2 \times 16 = 64 g/mol\) Mass of \(CO_2 = 0.02039 mol \times 44 g/mol = 0.8972 g\) Mass of \(SO_2 = 0.04078 mol \times 64 g/mol = 2.6099 g\) b. \(NaNO_3(s) \rightarrow NaNO_2(s) + O_2(g)\)

Balance the chemical equation

\(2NaNO_3(s) \rightarrow 2NaNO_2(s) + O_2(g)\)

Convert mass to moles for \(NaNO_3\)

Given mass of \(NaNO_3 = 1.55 g\). The molar mass of \(NaNO_3 = 23 + 14 + 3 \times 16 = 85 g/mol\). Therefore, moles of \(NaNO_3 = \frac{1.55}{85} = 0.01824 mol\)

Calculate moles of products

Using the balanced equation: \(2mol\; NaNO_3 \rightarrow 2mol\; NaNO_2\) and \(2mol\; NaNO_3 \rightarrow 1mol\; O_2\) Moles of \(NaNO_2 = 0.01824 mol\) Moles of \(O_2 = \frac{1}{2} \times 0.01824 mol = 0.00912 mol\)

Convert moles of products to grams

The molar mass of \(NaNO_2 = 23 + 14 + 2 \times 16 = 69 g/mol\) The molar mass of \(O_2 = 2 \times 16 = 32 g/mol\) Mass of \(NaNO_2 = 0.01824 mol \times 69 g/mol = 1.2586 g\) Mass of \(O_2 = 0.00912 mol \times 32 g/mol = 0.2918 g\) c. \(H_2(g) + MnO_2(s) \rightarrow MnO(s) + H_2O(g)\)

Balance the chemical equation

\(2H_2(g) + MnO_2(s) \rightarrow MnO(s) + 2H_2O(g)\)

Convert mass to moles for \(MnO_2\)

Given mass of \(MnO_2 = 1.55 g\). The molar mass of \(MnO_2 = 55 + 2 \times 16 = 87 g/mol\). Therefore, moles of \(MnO_2 = \frac{1.55}{87} = 0.01782 mol\)

Calculate moles of products

Using the balanced equation: \(1mol\; MnO_2 \rightarrow 1mol\; MnO\) and \(1mol\; MnO_2 \rightarrow 2mol\; H_2O\) Moles of \(MnO = 0.01782 mol\) Moles of \(H_2O = 2 \times 0.01782 mol = 0.03564 mol\)

Convert moles of products to grams

The molar mass of \(MnO = 55 + 16 = 71 g/mol\) The molar mass of \(H_2O = 2 \times 1 + 16 = 18 g/mol\) Mass of \(MnO = 0.01782 mol \times 71 g/mol = 1.2652 g\) Mass of \(H_2O = 0.03564 mol \times 18 g/mol = 0.6412 g\) d. \(Br_2(l) + Cl_2(g) \rightarrow BrCl(g)\)

Balance the chemical equation

\(Br_2(l) + 2Cl_2(g) \rightarrow 2BrCl(g)\)

Convert mass to moles for \(Br_2\)

Given mass of \(Br_2 = 1.55 g\). The molar mass of \(Br_2 = 2 \times 80 = 160 g/mol\). Therefore, moles of \(Br_2 = \frac{1.55}{160} = 0.00969 mol\)

Calculate moles of product

Using the balanced equation: \(1mol\; Br_2 \rightarrow 2mol\; BrCl\) Moles of \(BrCl = 2 \times 0.00969 mol = 0.01938 mol\)

Convert moles of product to grams

The molar mass of \(BrCl = 80 + 35.5 = 115.5 g/mol\) Mass of \(BrCl = 0.01938 mol \times 115.5 g/mol = 2.2379 g\)

Key concepts

Chemical Equations

Chemical equations are used to express the reactants and products in a chemical reaction. They represent how molecules interact and transform during reactions. A chemical equation includes the chemical formulas of reactants on the left side, an arrow in the middle indicating the direction of the reaction, and the products on the right side. For example, in the reaction

• \( ext{CS}_2(l) + ext{O}_2(g) ightarrow ext{CO}_2(g) + ext{SO}_2(g) \)

this indicates that carbon disulfide reacts with oxygen to form carbon dioxide and sulfur dioxide.

Writing chemical equations helps to visualize the changes that occur and forms the foundation for calculations used in stoichiometry. It is essential to properly write all substances involved with their correct chemical formulas to avoid any errors in calculations.

Molar Mass

The concept of molar mass links the mass of a substance to the number of moles, enabling us to perform stoichiometric calculations. Molar mass is defined as the mass of one mole of a substance, typically expressed in grams per mole (g/mol). To determine the molar mass of a compound, sum up the atomic masses of its constituent elements.
For instance,

• the molar mass of \( CS_2 \) is calculated by adding the atomic mass of one carbon atom (\( C = 12 \, ext{g/mol} \)) with that of two sulfur atoms (\( S = 32 \, ext{g/mol} \)), resulting in \( 76 \, ext{g/mol} \).

This information allows conversion from mass to moles, a crucial step in stoichiometry to find the moles of reactants and products in a chemical reaction. Understanding molar mass enables a deeper comprehension of the scales at which chemical reactions occur.

Chemical Reactions

Chemical reactions involve the transformation of reactants to products through the breaking and forming of chemical bonds. This process can be shown through a balanced chemical equation, which provides insight into the quantitative aspect of reactants and products involved.
For example, in a typical combustion reaction with carbon disulfide,

• \( ext{CS}_2(l) + 3 ext{O}_2(g) \rightarrow ext{CO}_2(g) + 2 ext{SO}_2(g) \), the elements within the reactants rearrange to form different products.

Reactants are consumed, and new substances with different properties are formed. Recognizing the nature of these transformations is key to predicting and understanding the behavior of materials during experiments.

Balancing Equations

Balancing equations is a critical step ensuring that the mass and the number of atoms are the same on both sides of a chemical equation. It reflects the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction.
Balancing requires adjusting coefficients in the chemical equation to ensure the same number of each atom type appears in the reactants and products.

• For example, to balance the equation \( ext{CS}_2(l) + ext{O}_2(g) \rightarrow ext{CO}_2(g) + ext{SO}_2(g) \), it becomes \( ext{CS}_2(l) + 3 ext{O}_2(g) \rightarrow ext{CO}_2(g) + 2 ext{SO}_2(g) \).

This ensures that there are as many oxygen atoms entering as reactants as emerging in the products. Balanced equations are essential for accurate stoichiometric calculations and understanding chemical transformations.