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3 years ago

Mastering Problems

Chemistry

1 answer:

KatRina [158]3 years ago

3 0

a. HI(g) ⇒ H₂9g) + I₂(g)

b. Al(s) + I₂(s)⇒AlI₃(s)

c. FeO(s)+O₂(g)⇒Fe₂O₃(s)

<h3>Further explanation</h3>

Given

word equations

Required

skeleton equations

Solution

The skeleton equation shows the chemical formula of the reaction between reactants and products, regardless of the actual amount of the reacting compound (unbalanced reaction)

The skeleton equations :

a. HI(g) ⇒ H₂9g) + I₂(g)

b. Al(s) + I₂(s)⇒AlI₃(s)

c. FeO(s)+O₂(g)⇒Fe₂O₃(s)

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In the equation below

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Answer:

6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

Explanation:

We are given the chemical equation:

$\displaystyle 2\text{NH$_3$}_\text{(g)} + 3\text{CuO}_\text{(s)} \longrightarrow \text{N$_2$}_\text{(g)} + 3\text{Cu}_\text{(s)}+3\text{H$_2$O}_\text{(g)}$

And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.

Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:

The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)

The ratio between NH₃ and Cu is 2:3.
The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)

Dimensional Analysis:

The amount of N₂ produced:

$\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{1\text{ mol N$_2$}}{2\text{ mol NH$_3$}} = 6.25\text{ mol N$_2$}$

The amount of Cu produced:

$\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol Cu}}{2\text{ mol NH$_3$}} = 18.8\text{ mol Cu}$

And the amount of H₂O produced:

$\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol H$_2$O}}{2\text{ mol NH$_3$}} = 18.8\text{ mol H$_2$O}$

In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

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Answer:

Explanation:

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Answer:

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Explanation:

Let's consider the following balanced equation.

2 Fe³⁺(aq) + Sn²⁺(aq) ⇒ 2Fe²⁺(aq) + Sn⁴⁺(aq)

First, we have to calculate the moles of Sn²⁺ that react.

$\frac{0.1015molSn^{2+} }{1L} .13.28 \times 10^{-3} L=1.348\times 10^{-3}molSn^{2+}$

We also know the following relations:

According to the balanced equation, 1 mole of Sn²⁺ reacts with 2 moles of Fe³⁺.
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Then, for 1.348 × 10⁻3 moles of Sn²⁺:

$1.348\times 10^{-3}molSn^{2+}.\frac{2molFe^{3+} }{1molSn^{2+} } .\frac{1molFe}{1molFe^{3+} } .\frac{55.84gFe}{1molFe} =0.1505gFe$

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