Chapter 27: Q22 P (page 768)

A mixture containing only ${Al}_{2} O_{3}$ (FM 101.96) and ${Fe}_{2} O_{3}$ (FM 159.69) weighs 2.019 g. When heated under a stream of $H_{2} {Al}_{2} O_{3}$ is unchanged, butis ${Fe}_{2} O_{3}$ converted into metallic Fe plus $H_{2} O (g)$ If the residue weighs 1.774 g, what is the weight percent of ${Fe}_{2} O_{3}$ in the original mixture?

Short Answer

Expert verified

The weight percent of ${Fe}_{2} O_{3}$ is 0.404%.

Step by step solution

Calculate the weight percent:

From the question we can consider the reaction as,

${Fe}_{2} O_{3} + {Al}_{2} O_{3} + heat 鈫� Fe + {Al}_{2} O_{3}$

Consider,

localid="1663665835168" $m_{1} ({Fe}_{2} O_{3} + {Al}_{2} O_{3})$ as 2.019g,

$m_{2} (Fe + {Al}_{2} O_{3})$ as 1.774g

Mass of lost oxygen is $m (O) = m_{1} - m_{2}$

$m (O) = 2.019 g - 1.774 g m (O) = 0.245 g 0.245 g o f o x y g e n = 0.01531 m o l e s$

Calculate the moles of localid="1663665958980" ${Fe}_{2} O_{3}$

$n ({Fe}_{2} O_{3}) = \frac{1}{3} 脳 n (O) n ({Fe}_{2} O_{3}) = \frac{1}{3} 脳 0.01531 mol n ({Fe}_{2} O_{3}) = 0.005105 mol$

So the weight percentage is

$w t % = \frac{m (F e_{2} O_{3})}{m (s a m p l e)} w t % = \frac{0.815 g}{2.019 g} w t % = 0.404$

The weight percent of $F e_{2} O_{3}$ is 0.404%.

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Short Answer

Step by step solution

Calculate the weight percent:

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A mixture containing only Al2O3(FM 101.96) andFe2O3(FM 159.69) weighs 2.019 g. When heated under a stream ofH2Al2O3is unchanged, butisFe2O3 converted into metallic Fe plusH2O(g)If the residue weighs 1.774 g, what is the weight percent ofFe2O3in the original mixture?

Short Answer

Step by step solution

Calculate the weight percent:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Chemistry Branches

Physical Chemistry

Ionic and Molecular Compounds

Inorganic Chemistry

The Earths Atmosphere

Study anywhere. Anytime. Across all devices.