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Moles Questions... Please Help If You Can

It would be great if anyone could help me with these questions:
Note: equation is Na2CO3+ 2HNO3>>>>2NaNO3+H20+CO2
1) If 0.56g of Na2CO3 was used what volume of CO2 would be produced at room temperature and pressure?

2) A student reacted 5.83g of Ba metal with H2O to form 500cm3 of Ba(OH)2
solution
a) Calculate the concentration of the resulting solution.

b) What volume of hydrogen gas would be produced?


Ps: I've worked them out but the answers don't seem right so please do show all your working out. Thanks in advance.
Original post by Angel May
It would be great if anyone could help me with these questions:
Note: equation is Na2CO3+ 2HNO3>>>>2NaNO3+H20+CO2
1) If 0.56g of Na2CO3 was used what volume of CO2 would be produced at room temperature and pressure?

2) A student reacted 5.83g of Ba metal with H2O to form 500cm3 of Ba(OH)2
solution
a) Calculate the concentration of the resulting solution.

b) What volume of hydrogen gas would be produced?


Ps: I've worked them out but the answers don't seem right so please do show all your working out. Thanks in advance.


If you show your working people will be able to point out where you are going wrong (or right).

It is unreasonable to expect folks to do your homework without some personal input...
Reply 2
Original post by charco
If you show your working people will be able to point out where you are going wrong (or right).

It is unreasonable to expect folks to do your homework without some personal input...

Oh wow well thanks but since when did it become a crime to ask for help?? And for the records I have done the questions and I didn't want to include the the answers in the previous post because they really don't seem right and to be honest the last thing I want is for a stranger to criticise me.

Ps: To anyone who is willing to help me see below for my working out.
Q1) n= mass\mr so I did 0.56/106 = 0.005 moles of Na2CO3
From the equation I figured that ratio of Na2CO3:CO2 is 1:1 therefore the moles must be the same for both. And to calculate the volume of gas produced, I used the formula v= n*24000 since the unit for the particular question is cm3.My final answer is 126.79.

Q2) Ba+2H2O »»» Ba(OH)2 +H2 (balanced)
5.83/137.3 =0.4246 moles of Ba
moles of Ba(OH)2 is the same as seen from ratio in the equation so I did 0.04246/0.5 ( note that I changed 500cm3 to dm3) and got an answer of 0.0849( doesn't seem right though).

b) It seems like I have to use the formula n=cv but I'm not entirely sure on that one.

Please do correct me if I'm wrong but be nice about it. Thanks in advance
Reply 3
Original post by Angel May
Oh wow well thanks but since when did it become a crime to ask for help?? And for the records I have done the questions and I didn't want to include the the answers in the previous post because they really don't seem right and to be honest the last thing I want is for a stranger to criticise me.

Ps: To anyone who is willing to help me see below for my working out.
Q1) n= mass\mr so I did 0.56/106 = 0.005 moles of Na2CO3
From the equation I figured that ratio of Na2CO3:CO2 is 1:1 therefore the moles must be the same for both. And to calculate the volume of gas produced, I used the formula v= n*24000 since the unit for the particular question is cm3.My final answer is 126.79.

Q2) Ba+2H2O »»» Ba(OH)2 +H2 (balanced)
5.83/137.3 =0.4246 moles of Ba
moles of Ba(OH)2 is the same as seen from ratio in the equation so I did 0.04246/0.5 ( note that I changed 500cm3 to dm3) and got an answer of 0.0849( doesn't seem right though).

b) It seems like I have to use the formula n=cv but I'm not entirely sure on that one.

Please do correct me if I'm wrong but be nice about it. Thanks in advance


I believe that Question 1 and 2 are correct.

for part b), you have to consider the molar ratio, and use moles=volume/24000(cm3)
Reply 4
Original post by Konflict
I believe that Question 1 and 2 are correct.

for part b), you have to consider the molar ratio, and use moles=volume/24000(cm3)


Oh really thanks. I wasn't sure on the formula for part b because it didn't state under room temperature and pressure in the question. I guess that was why I found it tricky but thanks to you it's not anymore:wink:. I really am grateful.

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