Chapter 4, Problem 1SAQ

Interpretation Introduction

To determine: The mass of Al required to completely react with $25.0\text{g}\text{M}\text{n}{\text{O}}_2$.

Answer to Problem 1SAQ

Solution: (d)

Explanation of Solution

Molar mass of $\text{M}\text{n}{\text{O}}_2=55+2\times 16=55+32=87$

Number of moles of $\text{M}\text{n}{\text{O}}_2=\frac{25.0}{87}=0.287$

Number of moles of $\text{A}\text{l}=0.287\times \left(\frac{4\text{m}\text{o}\text{l}\text{A}\text{l}}{3\text{m}\text{o}\text{l}\text{M}\text{n}{\text{O}}_2}\right)$

$=0.287\times \left(\frac{4\text{m}\text{o}\text{l}\text{A}\text{l}}{3\text{m}\text{o}\text{l}\text{M}\text{n}{\text{O}}_2}\right)$

= 0.382 mol Al

$\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{A}\text{l}=\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\times \text{M}\text{o}\text{l}\text{a}\text{r}\text{m}\text{a}\text{s}\text{s}\text{o}\text{f}\text{A}\text{l}=0.382\times 27=10.3\text{g}$.