Chapter 10.2, Problem 15PP

(a)

Interpretation Introduction

Interpretation:

The mass of 3.57 mol of Al needs to be determined.

Concept introduction:

Mass of an element in the product of the moles with the gram molecular weight of the respective element.

Answer to Problem 15PP

The mass of $3\cdot 57\text{ mol}$ of Al is $=96\cdot 318\text{ g}$.

Explanation of Solution

Given information:

Number of moles of $\text{Al = }3\cdot 57\text{ mol}$

Formula used:

$\text{weight = moles }\times \text{gram molecular weight}$

Calculation:

Given number of moles of Al is $3\cdot 57\text{ mol}$

Atomic weight of Al is $26\cdot 98$.

Substitute these in the above formula

$\begin{array}{l}\text{weight = moles }\times \text{gram molecular weight}\\ \text{=3}\cdot 57\times 26\cdot 98\\ =96\cdot 3186\end{array}$

Therefore, weight of Al is $96\cdot 3186\text{ g}$.

Interpretation Introduction

(b)

Interpretation:

The mass of Si needs to be determined.

Concept introduction:

Mass of an element in the product of the moles with the gram molecular weight of the respective element.

Answer to Problem 15PP

The mass of $42\cdot 6\text{ mol}$ of $\text{Si}=1192\cdot 8\text{ g}$

Explanation of Solution

Given information:

Number of moles of $\text{Si = }42\cdot 6$ mol.

Formula used:

$\text{weight = moles }\times \text{gram molecular weight}$

Calculation:

Given number of moles of $\text{Si = }42\cdot 6$

Atomic weight of $\text{Si = 28}$.

Substitute these in the above formula

$\begin{array}{l}\text{weight = moles }\times \text{gram molecular weight}\\ \text{=42}\cdot 6\times 28\\ =1192\cdot 8\text{ g}\end{array}$

Therefore, weight of Si is $=1192\cdot 8\text{ g}$.