Answered

Discover new information and get your questions answered with IDNLearn.com. Our experts provide prompt and accurate answers to help you make informed decisions on any topic.

Question 6:

Sodium thiosulphate, [tex]Na_2S_2O_3(s)[/tex], reacts with [tex]200 \, cm^3[/tex] of a hydrochloric acid solution, [tex]HCl(aq)[/tex], of concentration [tex]0.2 \, mol \cdot dm^{-3}[/tex] according to the following balanced equation:

[tex]\[ Na_2S_2O_3(s) + 2 HCl(aq) \rightarrow 2 NaCl(aq) + S(s) + SO_2(g) + H_2O(l) \][/tex]

6.1.1 Define the term concentration of a solution.

6.1.2 Calculate the number of moles of [tex]HCl(aq)[/tex] added to the sodium thiosulphate.

6.1.3 Calculate the volume of [tex]SO_2(g)[/tex] that will be formed if the reaction takes place at STP.

Sagot :

GinnyAnsweridn
Sure, let's go through the problem step by step.

### 6.1.1 Define the term concentration of a solution.

The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent or solution. It is commonly expressed in moles per liter (mol/L), which can also be referred to as molarity (M).

### 6.1.2 Calculate the number of moles of [tex]$HCl( aq )$[/tex] added to sodium thiosulphate.

We are given:
- The concentration of the hydrochloric acid solution, [tex]\( HCl \)[/tex], is [tex]\( 0.2 \, \text{mol/dm}^3 \)[/tex].
- The volume of the hydrochloric acid solution is [tex]\( 200 \, \text{cm}^3 \)[/tex].

First, we need to convert the volume from [tex]\(\text{cm}^3\)[/tex] to [tex]\(\text{dm}^3\)[/tex]:

[tex]\[ 1 \, \text{dm}^3 = 1000 \, \text{cm}^3 \][/tex]

[tex]\[ \text{Volume of } HCl = 200 \, \text{cm}^3 \times \frac{1 \, \text{dm}^3}{1000 \, \text{cm}^3} = 0.2 \, \text{dm}^3 \][/tex]

We can now calculate the number of moles of [tex]\( HCl \)[/tex]:

[tex]\[ \text{Number of moles of } HCl = \text{Concentration of } HCl \times \text{Volume of } HCl \][/tex]

[tex]\[ \text{Number of moles of } HCl = 0.2 \, \text{mol/dm}^3 \times 0.2 \, \text{dm}^3 = 0.04 \, \text{mol} \][/tex]

### Result:
The number of moles of [tex]\( HCl \)[/tex] added to the sodium thiosulphate is [tex]\( 0.04 \)[/tex] mol.

### 6.1.3 Calculate the volume of [tex]\(SO _2( g )\)[/tex] that will be formed if the reaction takes place at STP.

From the balanced equation, we know that [tex]\( 2 \)[/tex] moles of [tex]\( HCl \)[/tex] produce [tex]\( 1 \)[/tex] mole of [tex]\( SO_2 \)[/tex]:

[tex]\[ 2 \, HCl(aq) \rightarrow 1 \, SO_2(g) \][/tex]

So, the moles of [tex]\( SO_2 \)[/tex] produced can be calculated by:

[tex]\[ \text{Moles of } SO_2 = \frac{\text{Moles of } HCl}{2} \][/tex]

[tex]\[ \text{Moles of } SO_2 = \frac{0.04 \, \text{mol}}{2} = 0.02 \, \text{mol} \][/tex]

At Standard Temperature and Pressure (STP), 1 mole of a gas occupies [tex]\( 22.4 \, \text{dm}^3 \)[/tex]. Therefore, the volume of [tex]\( SO_2 \)[/tex] produced can be calculated by:

[tex]\[ \text{Volume of } SO_2 = \text{Moles of } SO_2 \times 22.4 \, \text{dm}^3/\text{mol} \][/tex]

[tex]\[ \text{Volume of } SO_2 = 0.02 \, \text{mol} \times 22.4 \, \text{dm}^3/\text{mol} = 0.448 \, \text{dm}^3 \][/tex]

To convert [tex]\(\text{dm}^3\)[/tex] to [tex]\(\text{cm}^3\)[/tex]:

[tex]\[ \text{Volume of } SO_2 = 0.448 \, \text{dm}^3 \times 1000 \, \text{cm}^3/\text{dm}^3 = 448 \, \text{cm}^3 \][/tex]

### Result:
The volume of [tex]\( SO_2 \)[/tex] gas formed at STP is [tex]\( 448 \, \text{cm}^3 \)[/tex].
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and see you next time for more reliable information.