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Chapter 3: Problem 187

Lanthanum was reacted with hydrogen in a given experiment to produce the nonstoichiometric compound \(\mathrm{LaH}_{2.90}\) Assuming that the compound contains \(\mathrm{H}^{-}, \mathrm{La}^{2+},\) and \(\mathrm{La}^{3+},\) calculate the fractions of \(\mathrm{La}^{2+}\) and \(\mathrm{La}^{3+}\) present.

Short Answer

Expert verified
The fractions of \(\mathrm{La}^{2+}\) and \(\mathrm{La}^{3+}\) ions present in the \(\mathrm{LaH}_{2.90}\) compound are: Fraction of \(\mathrm{La}^{2+}\): x = 0.10 Fraction of \(\mathrm{La}^{3+}\): y = 0.90 This means that 10% of the \(\mathrm{La}\) ions in the compound are in the \(\mathrm{La}^{2+}\) form and 90% are in the \(\mathrm{La}^{3+}\) form.

Step by step solution

01

1. Write the general formula for the compound

Considering that the compound contains \(\mathrm{H}^{-}\), \(\mathrm{La}^{2+}\), and \(\mathrm{La}^{3+}\), the general formula for the compound can be represented as follows: \[ \mathrm{La}^{2+}_{x}\mathrm{La}^{3+}_{y}\mathrm{H}^{-}_{z} \]
02

2. Write the charge balance equation

The charge balance equation states that the total positive charge must be equal to the total negative charge. In this case, the charges from \(\mathrm{La}^{2+}\) and \(\mathrm{La}^{3+}\) ions must be equal to the charge from the \(\mathrm{H}^{-}\) ions. So, we can write: \[ 2x + 3y = z \]
03

3. Write the composition constraint equation

The composition constraint equation relates the stoichiometry of the compound to the number of atoms of each ion in the whole compound. In this problem, the stoichiometry of the compound is \(\mathrm{LaH}_{2.90}\); therefore, we can write: \[ x + y = 1 \]
04

4. Solve the system of equations

Now let's solve the system of equations composed of the charge balance equation and the composition constraint equation to find the fractions of \(\mathrm{La}^{2+}\) and \(\mathrm{La}^{3+}\) ions. First, let's express the second equation in terms of x and substitute it into the first equation: From the second equation, we get: \[ x = 1 - y \] Substituting in the first equation, we get: \[ 2(1 - y) + 3y = 2.90 \] Solving this equation, we find y: \[ 3y-2y=2.90-2 \] \[ y=0.90 \] Now, substituting the value of y in the expression for x, we get: \[ x=1-y=1-0.90 \] \[ x=0.10 \]
05

5. Interpret the results

The fractions of \(\mathrm{La}^{2+}\) and \(\mathrm{La}^{3+}\) ions present in the \(\mathrm{LaH}_{2.90}\) compound are: Fraction of \(\mathrm{La}^{2+}\): x = 0.10 Fraction of \(\mathrm{La}^{3+}\): y = 0.90 This means that 10% of the \(\mathrm{La}\) ions in the compound are in the \(\mathrm{La}^{2+}\) form and 90% are in the \(\mathrm{La}^{3+}\) form.

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Most popular questions from this chapter

A sample of urea contains $1.121 \mathrm{g} \mathrm{N}, 0.161 \mathrm{g} \mathrm{H}, 0.480 \mathrm{g} \mathrm{C}\( and 0.640 \)\mathrm{g}$ O. What is the empirical formula of urea?

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