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When potassium cyanide ( \(\mathrm{KCN}\) ) reacts with acids, a deadly poisonous gas, hydrogen cyanide (HCN), is given off. Here is the equation: $$ \mathrm{KCN}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{KCl}(a q)+\mathrm{HCN}(g) $$ If a sample of \(0.140 \mathrm{~g}\) of \(\mathrm{KCN}\) is treated with an excess of \(\mathrm{HCl}\), calculate the amount of HCN formed, in grams.

Short Answer

Expert verified
The mass of HCN formed is approximately 0.0581 g.

Step by step solution

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01

Calculate the molar mass of KCN

First, determine the molar mass of KCN. The atomic masses are: K = 39.1, C = 12.0, and N = 14.0. Add these together to get the molar mass of KCN:\[39.1 + 12.0 + 14.0 = 65.1\text{ g/mol}\]
02

Convert grams of KCN to moles

Use the molar mass of KCN to convert 0.140 g of KCN to moles:\[\text{moles of KCN} = \frac{0.140}{65.1} \approx 0.00215 \text{ moles}\]
03

Use stoichiometry to find moles of HCN

From the balanced equation, the molar ratio of KCN to HCN is 1:1. Therefore, the moles of HCN produced is the same as the moles of KCN initially:\[\text{moles of HCN} = 0.00215 \text{ moles}\]
04

Calculate the mass of HCN formed

Determine the molar mass of HCN. The atomic masses are: H = 1.0, C = 12.0, and N = 14.0. Add these together:\[1.0 + 12.0 + 14.0 = 27.0 \text{ g/mol}\]Use the moles of HCN to find the grams of HCN:\[\text{mass of HCN} = 0.00215 \times 27.0 \approx 0.0581 \text{ g}\]

Key Concepts

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

Chemical Reactions
Chemical reactions are processes where substances, known as reactants, transform into different substances, called products. In the reaction given, potassium cyanide (KCN) reacts with hydrochloric acid (HCl) to produce potassium chloride (KCl) and hydrogen cyanide (HCN), a gas that has toxic properties.

In this process, atoms are rearranged but not created or destroyed. This particular reaction is an example of a single replacement reaction, where an element in a compound is replaced by an element in another compound.

Understanding the type of reaction occurring helps in predicting the products and the behavior of the substances involved. It also provides insight into the safety precautions needed, especially when dealing with dangerous chemicals.
Molar Mass Calculation
Molar mass is a fundamental concept in chemistry, referring to the mass of one mole of a given substance. It provides a bridge between the atomic scale and the bulk measurements in the laboratory.

To calculate the molar mass of \( \text{KCN} \), we sum the atomic masses of its constituent elements: potassium (K), carbon (C), and nitrogen (N).
  • Potassium: 39.1 g/mol
  • Carbon: 12.0 g/mol
  • Nitrogen: 14.0 g/mol
Adding these gives us a total molar mass: \( 39.1 + 12.0 + 14.0 = 65.1 \text{ g/mol} \). This value is crucial for converting between grams and moles, enabling precise stoichiometric calculations, which are essential for predicting how much of each product is formed in reactions.
Balanced Chemical Equations
Balanced chemical equations are vital for accurately representing chemical reactions. They ensure the conservation of mass, which means the same number of each type of atom is present in both the reactants and products.

In the equation \( \text{KCN} (aq) + \text{HCl} (aq) \rightarrow \text{KCl} (aq) + \text{HCN} (g) \), each molecule of KCN reacts with a molecule of HCl to produce one molecule each of KCl and HCN.

The balanced equation reflects a 1:1 molar ratio between the reactants and products, crucial for stoichiometric calculations. This ratio tells us that for every mole of KCN consumed, one mole of HCN is produced.
  • This ensures that the equation respects the law of conservation of mass and predicts yields of products.
  • Understanding this allows chemists to calculate the amounts of reactants needed or products formed and apply these skills effectively in laboratory settings or industry processes.
Balancing equations is not just an academic exercise; it’s an essential skill in chemistry used to ensure reactions proceed as desired.

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Most popular questions from this chapter

Determine the empirical formulas of the compounds with the following compositions: (a) 2.1 percent \(\mathrm{H}\), 65.3 percent \(\mathrm{O}, 32.6\) percent \(\mathrm{S} ;\) (b) 20.2 percent \(\mathrm{Al}\), 79.8 percent \(\mathrm{Cl}\).

Industrially, hydrogen gas can be prepared by combining propane gas \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) with steam at about \(400^{\circ} \mathrm{C}\). The products are carbon monoxide (CO) and hydrogen gas \(\left(\mathrm{H}_{2}\right) .\) (a) Write a balanced equation for the reaction. (b) How many kilograms of \(\mathrm{H}_{2}\) can be obtained from \(2.84 \times 10^{3} \mathrm{~kg}\) of propane?

The natural abundances of the two stable isotopes of hydrogen (hydrogen and deuterium) are 99.99 percent \({ }_{1}^{1} \mathrm{H}\) and 0.01 percent \({ }_{1}^{2} \mathrm{H}\). Assume that water exists as either \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{D}_{2} \mathrm{O} .\) Calculate the number of \(\mathrm{D}_{2} \mathrm{O}\) molecules in exactly \(400 \mathrm{~mL}\) of water \((\) density \(1.00 \mathrm{~g} / \mathrm{mL})\).

The depletion of ozone \(\left(\mathrm{O}_{3}\right)\) in the stratosphere has been a matter of great concern among scientists in recent years. It is believed that ozone can react with nitric oxide (NO) that is discharged from high-altitude jet planes. The reaction is $$ \mathrm{O}_{3}+\mathrm{NO} \longrightarrow \mathrm{O}_{2}+\mathrm{NO}_{2} $$ If \(0.740 \mathrm{~g}\) of \(\mathrm{O}_{3}\) reacts with \(0.670 \mathrm{~g}\) of NO, how many grams of \(\mathrm{NO}_{2}\) will be produced? Which compound is the limiting reactant? Calculate the number of moles of the excess reactant remaining at the end of the reaction.

Titanium(IV) oxide \(\left(\mathrm{TiO}_{2}\right)\) is a white substance produced by the action of sulfuric acid on the mineral ilmenite \(\left(\mathrm{FeTiO}_{3}\right):\) $$ \mathrm{FeTiO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{TiO}_{2}+\mathrm{FeSO}_{4}+\mathrm{H}_{2} \mathrm{O} $$ Its opaque and nontoxic properties make it suitable as a pigment in plastics and paints. In one process, \(8.00 \times\) \(10^{3} \mathrm{~kg}\) of \(\mathrm{FeTiO}_{3}\) yielded \(3.67 \times 10^{3} \mathrm{~kg}\) of \(\mathrm{TiO}_{2}\). What is the percent yield of the reaction?

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