91Ó°ÊÓ

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

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.

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

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

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

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

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

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

$=62.19\mathrm{g}$.

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

Now, we calculated given as below,

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

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

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

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

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

Now, we calculated as per question demand,

$=[2×223\mathrm{g}]\mathrm{PbO}→[2×239\mathrm{g}]\mathrm{PbS}$,

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

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

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