Question

I would like to be help with number 47 please

Answer 1

The percentage impurity is 64.875%

Step-by-step explanation

Given that;

The volume of the impure oxygen is 64 liters

The mass of water is 36.8 grams

Follow the steps below to find the percentage impurity

`\text{ 2H}_(2(g))\text{ + O}_(2(g))\text{ }\rightarrow\text{ 2H}_2O`

Step 1; Find the moles of water using the below formula

`\text{ mole = }\frac{\text{ mass }}{\text{ molar mass}}`

Recall, that the molar mass of water is 18.0 g/mol

`\begin{gathered} \text{ mole = }\frac{\text{ 36.8}}{\text{ 18.0}} \\ \text{ mole = 2.04 moles} \end{gathered}`

The moles of water is 2.04 moles

Step 2; Find the moles of oxygen using stoichiometry ratio

In the reaction, 1 mole of oxygen gives 2 moles of water

Let x represents the number of moles of oxygen

`\begin{gathered} \text{ 1 mole O}_2\text{ }\rightarrow\text{ 2 moles H}_2O \\ \text{ x moles O}_2\text{ }\rightarrow\text{ 2.04 moles H}_2O \\ \text{ cross multiply} \\ \text{ x moles O}_2\text{ }*\text{ 2 moles H}_2O\text{ }=\text{ 1 mole O}_2\text{ }*\text{ 2.04 moles H}_2O \\ \text{ Isolate x} \\ \text{ x = }\frac{1\text{ mole O}_2*2.04moles\cancel{H_2}O}{2moles\cancel{H_2}O} \\ \text{ x= }\frac{1\text{ }*\text{ 2.04}}{2} \\ \text{ x = 1.02 mole} \end{gathered}`

The mole of oxygen is 1.02 moles

Step 3; Find the volume of oxygen using a stoichiometry ratio

Recall, that 1 mole of O2 at STP is equivalent to 22.4 L/mol

Let x represents the volume of oxygen

`\begin{gathered} \text{ 1 mole O}_2\text{ }\rightarrow\text{ 22.4L/mol O}_2 \\ \text{ 1.02 moles O}_2\text{ }\rightarrow\text{ xL O}_2 \\ \text{ Cross multiply} \\ \text{ 1 mole O}_2\text{ }*\text{ xL O}_2\text{ }=\text{ 1.02 moles O}_2\text{ }*\text{ 22.4} \\ \text{ x = }\frac{1.02\text{ }*\text{ 22.4}}{1} \\ \text{ x = 22.848 L} \end{gathered}`

The volume of oxygen gas is 22.848L

Step 4; Find the percentage impurity of oxygen using the below formula

`\begin{gathered} \text{ \% impurity = }\frac{64\text{ - 22.848}}{64}*\text{ 100\%} \\ \\ \text{ \% impurity = }(41.152)/(64)*\text{ 100\%} \\ \text{ \% impurity = 0.64875 }*\text{ 100\%} \\ \text{ \% impurity = 64.875 \%} \end{gathered}`

Therefore, the percentage impurity is 64.875%