Struggling with two Chemistry questions
1. What is the volume occupied by 80cm3 of oxygen at r.t.p.
2. What mass of magnesium oxide (MgO) is produced from 72kg of magnesium? Write a balanced equation first.
I know the equation: concentration = amount (M) / volume (L). I don't know how to implement it into question 1. Question 2, writing a balanced equation I'm struggling with.
2. What mass of magnesium oxide (MgO) is produced from 72kg of magnesium? Write a balanced equation first.
I know the equation: concentration = amount (M) / volume (L). I don't know how to implement it into question 1. Question 2, writing a balanced equation I'm struggling with.
Original post by Mermeth
1. What is the volume occupied by 80cm3 of oxygen at r.t.p.
2. What mass of magnesium oxide (MgO) is produced from 72kg of magnesium? Write a balanced equation first.
I know the equation: concentration = amount (M) / volume (L). I don't know how to implement it into question 1. Question 2, writing a balanced equation I'm struggling with.
2. What mass of magnesium oxide (MgO) is produced from 72kg of magnesium? Write a balanced equation first.
I know the equation: concentration = amount (M) / volume (L). I don't know how to implement it into question 1. Question 2, writing a balanced equation I'm struggling with.
Whats rtp?
2Mg+O2----->2MgO
Is this for AS level? It looks like it might be.
1) You already have the volume. 80cm^3 is a volume.
2) Balanced equation:
2Mg + O2 ----> 2MgO
To get that you needed to write MgO first on the right hand side. You already know you need Mg on the left so write that too. Then the only element you're missing on the left that is on the right is O, which is always in diatomic form for these type of questions, so it's got to be O2.
Now you have more O on the left than you do on the right, so stick a 2 in front of the MgO so there are two oxygens on each side. Finally stick another 2 in front of the Mg on the left hand side as you now have 2 Mg on the right.
Does that help a bit?
*EDIT* Just occurred to me that this is probably a little basic for AS, apologies
1) You already have the volume. 80cm^3 is a volume.
2) Balanced equation:
2Mg + O2 ----> 2MgO
To get that you needed to write MgO first on the right hand side. You already know you need Mg on the left so write that too. Then the only element you're missing on the left that is on the right is O, which is always in diatomic form for these type of questions, so it's got to be O2.
Now you have more O on the left than you do on the right, so stick a 2 in front of the MgO so there are two oxygens on each side. Finally stick another 2 in front of the Mg on the left hand side as you now have 2 Mg on the right.
Does that help a bit?
*EDIT* Just occurred to me that this is probably a little basic for AS, apologies
(edited 9 years ago)
Original post by Nitrogen
Whats rtp?
Room Temperature and Pressure...
Original post by Dylann
Room Temperature and Pressure...
Because I've never came across this abbreviation.
Original post by raineandfyre
...you already have the volume. 80cm^3 is a volume.
That is what I thought. I think my teacher made a mistake with this question.
Original post by Mermeth
That is what I thought. I think my teacher made a mistake with this question.
Ok, well let me ask you a new question then
How many moles of Oxygen would occupy 80cm^3?
Original post by Mermeth
That is what I thought. I think my teacher made a mistake with this question.
Did they want the number of moles of gas at rtp?
If so you might need to use PV=nRT rearranged to give V=(nRT)/P, convert cm^3 to m^3, plug in values in Pa and K for standard pressure and temperature, and stick the numbers in a calculator.
Original post by raineandfyre
Did they want the number of moles of gas at rtp?
If so you might need to use PV=nRT rearranged to give V=(nRT)/P, convert cm^3 to m^3, plug in values in Pa and K for standard pressure and temperature, and stick the numbers in a calculator.
If so you might need to use PV=nRT rearranged to give V=(nRT)/P, convert cm^3 to m^3, plug in values in Pa and K for standard pressure and temperature, and stick the numbers in a calculator.
surely it would be n=PV/RT since n=number of moles?
Original post by Dylann
surely it would be n=PV/RT since n=number of moles?
Yes, you're right. This is what I get for trying to do things when I'm tired
Original post by raineandfyre
Yes, you're right. This is what I get for trying to do things when I'm tired
Ah don't worry, I find that chemistry is the most difficult subject to do when you're tired...I tell myself to not do any chemistry after about 7pm because I get everything wrong :P
Original post by Dylann
Ah don't worry, I find that chemistry is the most difficult subject to do when you're tired...I tell myself to not do any chemistry after about 7pm because I get everything wrong :P
That's a good idea... It never happens like that though.
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