Q. 7.65 - An Introduction to General, Organic, and Biological Chemistry

Question

When solid lead(II) sulfide reacts with oxygen gas, the products are solid lead(II) oxide and sulfur dioxide gas.

a. Write the balanced chemical equation for the reaction.

b. How many grams of oxygen are required to react with $29.9 \mathrm{g}$of lead(II) sulfide?

c. How many grams of sulfur dioxide can be produced when $65.0 \mathrm{g}$of lead(II) sulfide reacts?

d. How many grams of lead(II) sulfide are used to produce$128 \mathrm{g}$ of lead(II) oxide?

Step-by-Step Solution

Answer

(a) The balanced chemical equation is $2PbS(∆)+3O_2\left(g\right)\to 2PbO+2S{O}_2\left(g\right)$.

(b) The $62.19 \mathrm{g}$of oxygen are required to react with $29.9 \mathrm{g}$of lead(II) sulfide.

(c) The $16.9 \mathrm{g}$of sulfur dioxide can be produced when $65.0 \mathrm{g}$of lead(II) sulfide reacts.

(d) The $136.96 \mathrm{g}$of lead(II) sulfide are used to produce $128 \mathrm{g}$of lead(II) oxide.

Part(a) Step 1: Given information

We have been given, solid lead(II) sulfide reacts with oxygen gas, the products are solid lead(II) oxide and sulfur dioxide gas. 

Part(a) Step 2: Explanation

Now, we write balanced equation of given condition,

According to the procedure, solid lead to oxide and Sulphur dioxide are produced, so lead two sulfide (lead with a two plus charge and sulfide with a one negative charge) provides us one of each reacting with oxygen. Because oxygen is diatonic, it develops led to oxide, which has a charge of two plus.

$2PbS(∆)+3O_2\left(g\right)\to 2PbO+2S{O}_2\left(g\right)$

Part(b) Step 1: Given information

We have to find out the how many grams of oxygen are required to react with $29.9 \mathrm{g}$of lead(II) sulfide.

Part(b) Step 2: Explanation

Now, we calculate oxygen are required to react with $29.9 \mathrm{g}$of lead(II) sulfide,

$=[2\times 239 \mathrm{g}] \mathrm{PbS} [3\times 32 \mathrm{g}]{\mathrm{O}}_2$

$=[1 \mathrm{g}]\mathrm{PbS}\to \frac{3\times 32 \cancel{\mathrm{g}}}{2\times 239\cancel{ \mathrm{g}}}$

$=[29.9 \mathrm{g}]\mathrm{PbS}\to \frac{3\times 32}{2\times 239}\times 29.9 \mathrm{g} {\mathrm{O}}_2$

$=62.19 \mathrm{g}$.

Part(c) Step 1: Given information

We have to find out the how many  grams of sulfur dioxide can be produced when $38.5 \mathrm{g}$of oxygen reacts.

Part(c) Step 2: Explanation

Now, we calculated given as below,

$=[2\times 39 \mathrm{g}] \mathrm{PbS}\to [2\times 64 \mathrm{g}]{\mathrm{SO}}_2$,

$=\left[1\mathrm{g}\right] \mathrm{PbS}\to \frac{[2\times 64] \cancel{\mathrm{g}}}{[2\times 239] \cancel{\mathrm{g}}} {\mathrm{SO}}_2$,

$=[65 \mathrm{g}] \mathrm{PbS}\to \frac{[2\times 64]}{[2\times 239]}\times 65 \mathrm{g} {\mathrm{SO}}_2$,

$=16.9 \mathrm{g}$$S{O}_2$.

Part(d) Step 1: Given information

We have to find out the  how many grams of lead(II) sulfide are used to produce $55.8g$of water vapor.

Part(d) Step 2: Explanation

Now, we calculated as per question demand,

$=[2\times 223 \mathrm{g}] \mathrm{PbO}\to [2\times 239 \mathrm{g}] \mathrm{PbS}$,

$=[1 \mathrm{g}] \mathrm{PbO}\to \left[\frac{2\times 239 \cancel{\mathrm{g}}}{2\times 223 \cancel{\mathrm{g}}}\right]\mathrm{PbS}$,

$=[128 \mathrm{g}] \mathrm{PbO}\to \left[\frac{239}{223}\times 128 \mathrm{g}\right] \mathrm{PbS}$,

$=136.96 \mathrm{g} \mathrm{PbS}$.