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Chapter 19: Problem 40

Tin forms compounds in the +2 and +4 oxidation states. Therefore, when tin reacts with fluorine, two products are possible. Write balanced equations for the production of the two tin halide compounds and name them.

Short Answer

Expert verified
The balanced equations for the production of the two tin halide compounds are: 1. 2 Sn + F₂ → 2 SnF₂ (Tin(II) fluoride) 2. Sn + 2 F₂ → SnF₄ (Tin(IV) fluoride)

Step by step solution

01

Identify the tin halides

Tin can form two compounds with fluorine by reacting in the +2 and +4 oxidation states. These compounds are: - Tin(II) fluoride (SnF₂): In this compound, tin has an oxidation state of +2. - Tin(IV) fluoride (SnF₄): In this compound, tin has an oxidation state of +4.
02

Write the unbalanced equations

Now, let's write the unbalanced equations for the production of these tin halides: 1. Tin + Fluorine → Tin(II) fluoride: Sn + F₂ → SnF₂ 2. Tin + Fluorine → Tin(IV) fluoride: Sn + F₂ → SnF₄
03

Balance the equations

Now, we will balance both of these equations by adding coefficients to ensure that the number of atoms on each side of the equation is equal: 1. Sn + F₂ → SnF₂: To balance this equation, we simply need to multiply the F₂ by 1/2 since there is only 1 F₂ molecule in SnF₂. Balanced equation: Sn + 1/2 F₂ → SnF₂ 2. Sn + F₂ → SnF₄: To balance this equation, we need to multiply the F₂ by 2 since there are 4 fluorine atoms in SnF₄. Balanced equation: Sn + 2 F₂ → SnF₄
04

Rewrite the balanced equations with integers

It is sometimes preferred to write balanced chemical equations with integer coefficients. To accomplish this, we can multiply the coefficients by an appropriate factor to get rid of the fractions: 1. For Sn + 1/2 F₂ → SnF₂, we can multiply all coefficients by 2: Balanced equation with integers: 2 Sn + F₂ → 2 SnF₂ 2. For the second equation, the coefficients are already integers: Balanced equation with integers: Sn + 2 F₂ → SnF₄ In conclusion, we have written balanced equations for the production of two tin halide compounds: - 2 Sn + F₂ → 2 SnF₂ (Tin(II) fluoride) - Sn + 2 F₂ → SnF₄ (Tin(IV) fluoride)

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