Problem 86 - Introductory Chemistry Atoms First

Question

Hydrogen gas is produced when elemental tin reacts with HF to produce hydrogen and tin(II) fluoride. Write the balanced equation for this reaction.

Step-by-Step Solution

Answer

The balanced equation for the reaction between elemental tin and hydrogen fluoride to produce hydrogen gas and tin(II) fluoride is: \( Sn + 2HF → 2H2 + SnF2 \)

Step 1: 1. Write the unbalanced chemical equation

First, we need to write the unbalanced chemical equation using the correct chemical formulas for the reactants and the products. The reactants are elemental tin (Sn) and hydrogen fluoride (HF), and the products are hydrogen gas (H2) and tin(II) fluoride (SnF2). The unbalanced equation is: Sn + HF → H2 + SnF2

Step 2: 2. Identify the atoms to balance

Next, we need to identify the atoms that need to be balanced in the equation. In this case, those are the tin (Sn), hydrogen (H), and fluorine (F) atoms.

Step 3: 3. Balance the tin atoms (Sn)

There is one tin atom on the reactant side and one tin atom on the product side, so the tin atoms are already balanced.

Step 4: 4. Balance the hydrogen atoms (H)

On the reactant side, there are two hydrogen atoms in HF (because the balance of the fluorine atoms is not yet balanced). On the product side, there are two hydrogen atoms in H2. Therefore, for balancing the hydrogen atoms, we don't need to do anything, as the hydrogen atoms are already balanced for now.

Step 5: 5. Balance the fluorine atoms (F)

On the reactant side, there is one fluorine atom in HF and on the product side, there are two fluorine atoms in SnF2. So, we need to add a coefficient of 2 in front of the HF in the reactants to ensure there are as many fluorine atoms as in the products: Sn + 2HF → H2 + SnF2

Step 6: 6. Check the balance of hydrogen atoms (H) again

Now that we have balanced the fluorine atoms, we need to check if the hydrogen atoms are still balanced. On the reactant side, there are now four hydrogen atoms (2 moles of HF have 2 × 2 = 4 hydrogen atoms), and on the product side, there are two hydrogen atoms in H2. To balance the equation, put a coefficient of 2 in front of the H2 on the product side. Sn + 2HF → 2H2 + SnF2

Step 7: 7. Verify the balanced equation

Finally, let's verify that the balanced equation is correct. For each type of atom, count the number on both sides of the equation. Tin: 1 on both sides; Hydrogen: 4 on both sides (2 × 2 = 4); Fluorine: 2 on both sides. Thus, the balanced chemical equation for the reaction between elemental tin and hydrogen fluoride to produce hydrogen gas and tin(II) fluoride is: Sn + 2HF → 2H2 + SnF2

Key Concepts

Chemical Reactions

Chemical reactions involve the transformation of reactants into products through the rearrangement of atoms. In this exercise, we explore a chemical reaction where elemental tin reacts with hydrogen fluoride (HF) to produce hydrogen gas (H2) and tin(II) fluoride (SnF2). The unbalanced reaction can be written as:

• Reactants: Tin (Sn), Hydrogen fluoride (HF)
• Products: Hydrogen gas (H₂), Tin(II) fluoride (SnF₂)

Balancing this reaction requires understanding the role each element plays and ensuring the conservation of atoms of each element. Balancing chemical equations is crucial in stoichiometry, as it allows us to accurately describe how much of each substance is consumed and produced.

Stoichiometry

Stoichiometry is the study of quantitative relationships in chemical reactions. It involves calculating the amount of reactants required and the amount of products formed. For stoichiometry, balanced chemical equations are essential, as they provide the mole ratio of reactants and products. In the tin and hydrogen fluoride reaction:

• One mole of tin reacts with two moles of HF
• Yielding two moles of hydrogen gas and one mole of tin(II) fluoride

The balanced equation: \[\text{Sn + 2HF} \rightarrow \text{2H}_2 + \text{SnF}_2\]This equation indicates that the amount of hydrogen gas and tin(II) fluoride produced is directly proportional to the amounts of tin and hydrogen fluoride used.

Tin Chemistry

Tin is a versatile metal with the symbol Sn and is part of the carbon group on the periodic table. In chemistry, tin can form multiple compounds, such as tin(II) fluoride, which is one of the products in the reaction we are considering. When tin reacts with hydrogen fluoride, it undergoes an oxidation-reduction reaction:

• Tin (Sn) is oxidized from 0 to +2 oxidation state in tin(II) fluoride
• Meanwhile, the fluoride ions in HF facilitate the transfer of electrons

This redox behavior of tin demonstrates its ability to form stable compounds, making it an interesting element in chemistry.

Hydrogen Gas Production

Hydrogen gas production is a vital process in numerous chemical reactions and industrial applications. In this exercise, hydrogen gas is produced when tin reacts with hydrogen fluoride. Hydrogen gas (\[\text{H}_2\]) is highly reactive and plays a key role in various chemical processes:

• Used as a reducing agent
• Involved in the production of ammonia and methanol
• Contributes to hydrogenation processes in industries

The reaction showcasing hydrogen gas production reflects the efficient release of hydrogen atoms from hydrogen fluoride, which combines to form molecular hydrogen. This highlights the importance of understanding chemical equations and stoichiometry in determining the feasibility of producing useful chemical gases.