Question

For each of the following unbalanced chemical equations, suppose that exactly \(1.00 \mathrm{~g}\) of each reactant is taken. Determine which reactant is limiting, and calculate what mass of the product in boldface is expected (assuming that the limiting reactant is completely consumed). a. \(\mathrm{CS}_{2}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{SO}_{2}(g)\) b. \(\mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CN}_{2} \mathrm{H}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{MnO}_{2}(s) \rightarrow \mathrm{MnO}(s)+\mathbf{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{I}_{2}(l)+\mathrm{Cl}_{2}(g) \rightarrow \mathbf{I C l}(g)\)

Short answer

The expected masses of the products are: a. 1.33 g of SO₂ b. 1.09 g of CN₂H₄O c. 0.414 g of H₂O d. 1.28 g of ICl

Step-by-step solution

Balance the chemical equation

The balanced chemical equation is: \[CS_{2}(l) + 3O_{2}(g) \rightarrow CO_{2}(g) + 2SO_{2}(g)\]

Convert masses of reactants to moles

For 1.00 g of each reactant, determine the number of moles. Moles of CS₂: \(\frac{1.00 \ g}{76.14 \frac{g}{mol}} = 0.0131 \ mol\) Moles of O₂: \(\frac{1.00 \ g}{32.00 \frac{g}{mol}} = 0.0312 \ mol\)

Determine the limiting reactant

Compare the moles of reactants to the stoichiometry of the balanced equation: \(\frac{0.0131 \ mol \ CS_2}{1} \ : \frac{0.0312 \ mol \ O_2}{3}\) The moles of O₂ available would require 0.0104 moles of CS₂, but we only have 0.0131 moles. So, the limiting reactant is O₂.

Calculate the expected mass of desired product

We are looking for the mass of SO₂ produced. Use stoichiometry: Moles of SO₂ produced = \(\frac{2 \ mol \ SO_2}{3 \ mol \ O_2} \times 0.0312\,mol\,O_2 = 0.0208\,mol\, SO_2 \) Mass of SO₂ produced = \(0.0208\,mol\, SO_2 \times 64.07\frac{g}{mol} = 1.33\,g\,SO_2\) ##b. NH3(g) + CO2(g) -> CN2H4O(s) + H2O(g)##

Balance the chemical equation

The balanced chemical equation is: \[2NH_{3}(g) + CO_{2}(g) \rightarrow CN_{2}H_{4}O(s) + 2H_{2}O(g)\]

Convert masses of reactants to moles

For 1.00 g of each reactant, determine the number of moles. Moles of NH₃: \(\frac{1.00 \ g}{17.03 \frac{g}{mol}} = 0.0587 \ mol\) Moles of CO₂: \(\frac{1.00 \ g}{44.01 \frac{g}{mol}} = 0.0227 \ mol\)

Determine the limiting reactant

Compare the moles of reactants to the stoichiometry of the balanced equation: \(\frac{0.0587 \ mol \ NH_3}{2} : \frac{0.0227 \ mol \ CO_2}{1}\) Since 0.0293 moles of NH₃ are required for 0.0227 moles of CO₂, we have enough NH₃, so CO₂ is the limiting reactant.

Calculate the expected mass of desired product

We are looking for the mass of CN₂H₄O produced. Use stoichiometry: Moles of CN₂H₄O produced = \(\frac{1 \ mol \ CN_2H_4O}{1 \ mol \ CO_2} \times 0.0227 \ mol \ CO_2 = 0.0227\,mol\,CN_2H_4O \) Mass of CN₂H₄O produced = \(0.0227\,mol\, CN_2H_4O \times 48.05\frac{g}{mol} = 1.09\,g\,CN_2H_4O\) ##c. H2(g) + MnO2(s) -> MnO(s) + H2O(g)##

Balance the chemical equation

The balanced chemical equation is: \[2H_{2}(g) + MnO_{2}(s) \rightarrow MnO(s) + 2H_{2}O(g)\]

Convert masses of reactants to moles

For 1.00 g of each reactant, determine the number of moles. Moles of H₂: \(\frac{1.00 \ g}{2.02 \frac{g}{mol}} = 0.495 \ mol\) Moles of MnO₂: \(\frac{1.00 \ g}{86.94 \frac{g}{mol}} = 0.0115 \ mol\)

Determine the limiting reactant

Compare the moles of reactants to the stoichiometry of the balanced equation: \(\frac{0.495 \ mol \ H_2}{2} : \frac{0.0115 \ mol \ MnO_2}{1}\) Since 0.0229 moles of H₂ are required for 0.0115 moles of MnO₂, we have enough H₂, so MnO₂ is the limiting reactant.

Calculate the expected mass of desired product

We are looking for the mass of H₂O produced. Use stoichiometry: Moles of H₂O produced = \(\frac{2 \ mol \ H_2O}{1 \ mol \ MnO_2} \times 0.0115 \ mol \ MnO_2 = 0.0230\,mol\,H_2O \) Mass of H₂O produced = \(0.0230\,mol\, H_2O \times 18.02\frac{g}{mol} = 0.414\,g\,H_2O\) ##d. I2(l) + Cl2(g) -> ICl(g)##

Balance the chemical equation

The balanced chemical equation is: \[I_{2}(l) + Cl_{2}(g) \rightarrow 2ICl(g)\]

Convert masses of reactants to moles

For 1.00 g of each reactant, determine the number of moles. Moles of I₂: \(\frac{1.00 \ g}{253.81 \frac{g}{mol}} = 0.00394 \ mol\) Moles of Cl₂: \(\frac{1.00 \ g}{70.91 \frac{g}{mol}} = 0.0141 \ mol\)

Determine the limiting reactant

Compare the moles of reactants to the stoichiometry of the balanced equation: \(\frac{0.00394 \ mol \ I_2}{1} : \frac{0.0141 \ mol \ Cl_2}{1}\) We have more Cl₂ than required, so I₂ is the limiting reactant.

Calculate the expected mass of desired product

We are looking for the mass of ICl produced. Use stoichiometry: Moles of ICl produced = \(\frac{2 \ mol \ ICl}{1 \ mol \ I_2} \times 0.00394 \ mol \ I_2 = 0.00788\,mol\,ICl \) Mass of ICl produced = \(0.00788\,mol\, ICl \times 162.36\frac{g}{mol} = 1.28\,g\,ICl\)

Key concepts

Limiting Reactant

The concept of a limiting reactant is fundamental to chemical reactions, as it determines the maximum amount of product that can be formed. When different reactants are combined, often one of them will be used up before the others, stopping the reaction from proceeding further. This reactant is known as the limiting reactant because it limits the amount of products formed.

To identify the limiting reactant, one needs to perform a mole comparison based on the balanced chemical equation. The molar ratio of reactants, as given by the coefficients in the balanced equation, provides the exact proportion in which reactants combine. When we compare the available moles of reactant to these ratios, the reactant that provides the lesser amount of product, as dictated by the stoichiometry of the reaction, is the limiting reactant.

For instance, in the example where carbon disulfide reacts with oxygen to produce carbon dioxide and sulfur dioxide, after balancing the equation and converting the mass of each reactant to moles, we compare the mole ratio. Based on the amount and the stoichiometry within the balanced equation, oxygen is the limiting reactant, thereby defining the extent of the reaction and the mass of sulfur dioxide that can be produced.

Mole-to-Mass Conversion

Mole-to-mass conversion is a pivotal step in stoichiometry that allows one to relate the quantitative information in the chemical equation to measurable quantities. To perform this conversion, one must use the molar mass of the substance, which is the mass in grams of one mole of the substance.

The process is straightforward: divide the mass of the substance by its molar mass to obtain the number of moles. This step directly connects the conceptual world of moles and chemical equations to the practical world of grams and laboratory measurements. For example, if we start with 1.00 gram of hydrogen gas, we divide this mass by the molar mass of hydrogen (approximately 2.02 grams per mole) to obtain the number of moles of hydrogen gas available for the reaction with manganese dioxide. Similarly, this conversion is essential in calculating the mass of the product formed from the limiting reactant, as seen in the provided solutions where the mass of sulfur dioxide, urea, water, and iodine monochloride were calculated.

Balanced Chemical Equations

Balanced chemical equations are the skeleton of chemical stoichiometry. They are expressions which convey the proportional relationships between reactants and products in a chemical reaction, adhering to the law of conservation of mass. In a balanced equation, the number of atoms of each element on the reactant side must be equal to the number on the product side.

Take for example the reaction of ammonia with carbon dioxide. The equation can be balanced by adjusting the coefficients to ensure that the same number of nitrogen, hydrogen, carbon, and oxygen atoms appear on both sides. This balancing act is crucial as it dictates the molar ratios needed for mole-to-mass conversions and for determining the limiting reactant. The balanced equation serves as the basis for all stoichiometric calculations, allowing us to predict the quantities of substances consumed and produced in a chemical reaction.