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# Measuring volumes of gases watch

1. Hiya

This question sheet has no answers, so I can't tell if I'm right,
can anyone help?

MgCO3 + 2HCl -> MgCl2 + CO2 + H2O
0.800g of MgCO3 is added to 50 cm3 of 0.124moldm-3 HCl

a) What is the quantity, in mol, of MgCO3 added
b) What is the quantity, in mol, of HCl in 50cm3
c) What volume of CO2 will be produced
d) How much excess MgCO3 is added?
e) What would occur to the volume of CO2 if excess MgCO3 is reacted with ...
i) 0.124mol of Nitric acid is used (Increase/Decrease/Stay the same)
ii) 0.124mol of Sulfuric acid is used (Incrase/Decrease/Stay the same)

All to 3 sig fig
a) 0.00949 mol
b) 0.00620 mol
c) 0.129 cm3
d) 0.539g
e)
i) Stay the same
ii) Increase

gaby
2. (Original post by gaby_xo)
Hiya

This question sheet has no answers, so I can't tell if I'm right,
can anyone help?

MgCO3 + 2HCl -> MgCl2 + CO2 + H2O
0.800g of MgCO3 is added to 50 cm3 of 0.124moldm-3 HCl

a) What is the quantity, in mol, of MgCO3 added
b) What is the quantity, in mol, of HCl in 50cm3
c) What volume of CO2 will be produced
d) How much excess MgCO3 is added?
e) What would occur to the volume of CO2 if excess MgCO3 is reacted with ...
i) 0.124mol of Nitric acid is used (Increase/Decrease/Stay the same)
ii) 0.124mol of Sulfuric acid is used (Incrase/Decrease/Stay the same)

All to 3 sig fig
a) 0.00949 mol
b) 0.00620 mol
c) 0.129 cm3
d) 0.539g
e)
i) Stay the same
ii) Increase

gaby

For a), you can calculate the relative mass of MgCO3 and divide this by the mass to give you the amount of moles.

Once you have the formulae cracked it's a piece of cake.
3. (Original post by Veqz)

For a), you can calculate the relative mass of MgCO3 and divide this by the mass to give you the amount of moles.

Once you have the formulae cracked it's a piece of cake.
Thank you:-)
I am using the triangles, i'm just hoping to check my answers
:
4. (Original post by gaby_xo)
Thank you:-)
I am using the triangles, i'm just hoping to check my answers
:
Ah I see. Well if you need any more help then just let me know
6. (Original post by Borek)

(Original post by Veqz)
Ah I see. Well if you need any more help then just let me know
Thanks for your help, My CO2 was off when I looked again!
much thanks
G

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Updated: November 24, 2013
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