Can someone show me how these are done?

Question

What volume of O2 at 298K and 2 atm is required to react completely with 2L of Methane at the same temperature and pressure?

How much 0.5 M KCl solution is required to react completely with 100 mL of 2M AgNO3 solution? KCl + AgNO3 -> AgCl + KNO3

The points go to whoever responds first with correct answers and adequate work...

Wow, thanks Abe.

Answer 1

For the Methane combustion question you first need to look at the balanced equation.

You know that Methane is CH4 and it burns in Oxygen to form Carbon Dioxide and Water, lets look first at the unbalanced equation.

x CH4 + y 02 = w C02 + z H20

if you set X=1 and w=1 then you have the carbons already balanced.

You know that there are a total of 4 hydrogen atoms on the left side, this will reqire a 2 molecules of water on the right so the equation becomes

CH4 + yO2 = C02 + 2H20

Count the Oxygen atoms on the right to figure out how many Oxygen molecule you will need on the left. There are 4 Oxygen atoms on the right so you will need 2 Oxygen molecules on the left.

CH4 + 202 = C02 + 2H20

This equation is also true in terms of moles .

We know each mole at Standard Temp of 273 K and pressure of 1 atm has the volume of 22.4 Liters.

We need to know how many moles of Methane we are given, so we can figure out the amount of oxygen we will need.

remember the gas equation: pv = nrt

In this case the pressure is in atm and the volume is in liters, so to find the gas constant r (if you did not already know it)we divide pv/nt to get
r = 0.082 atm liters/mole K

Now n = pv/rt = 2atm*2Liters/(0.082 atm liters/mole K)*298
n (the number of moles of Methane) = 0.164

You will need twice as many moles of Oxygen molecules, so that would be 0.328 moles of Oxygen.

Note that the question was given in Liters, and since the same temperature and pressure was used for each gas (that we pretend is ideal) then all this extra work was really not needed.

Two liters of Methane would require 4 Liters of Oxygen.

If we did not make a mistake then 0.328 moles of Oxygen at 2 atm and 298K should take up 4 liters of volume, is this true?

pv = nrt -------> V = (nrt)/p = 0.328*0.082*298/2 = 4, yes it is.

****************
100ml* 1 Liter/1000ml * 2 Moles/Liter AgNO3 = 0.2 moles AgNO3

So, you will need in total 0.2 moles of KCL, but the question was asked in terms of volume of 0.5 M KCL.

0.5 Moles KCL/1000ml * X ml = 0.2 Moles of KCL

You will need 400 ml of 0.5 Molar KCL

Answer 2

consult a chem book. its difficult to answer that question here

Answer 3

am sorry i donnot know that no

Answer 4

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