The Student Room Group

Reply 1

Remember no mass can be lost or gained throughout the reaction. With this in mind we can find out the mass of the water on the right side of the equation:
4.92 - 2.40 = 2.52.

The molar mass of water (H2O) is 18 (1+1+16) so to find the number of moles of it you do the molar mass divided by the actual mass:

18.00 / 2.52 = 7.14

The value will never be perfect because of experimental error, so round it to 7. So there are 7 moles of water on the right side of the equation, so n = 7.

Hope that helps

Reply 2

Tomsey mass/Mr= moles :p:
Coincidentally, from my quick calculation, your answer is correct!

OP, work out the moles of water then the moles of magnesium sulphate. Then work out howmany waters to 1 mole of MgSO4, and this is n.

eg if you get 0.25 moles MgSO4 and 0.5 water, this works out 1:2. Methodically, divide by the smaller number to work out the ratio as you often get long numbers.

Reply 3

Original post
by squishedbanana
I really hope this isn't a hopelessly easy question and I'm just stupid. I just can't see how to go forward.:confused:

Reply 4

Original post
by silent ninja
Tomsey mass/Mr= moles :p:
Coincidentally, from my quick calculation, your answer is correct!

OP, work out the moles of water then the moles of magnesium sulphate. Then work out howmany waters to 1 mole of MgSO4, and this is n.

eg if you get 0.25 moles MgSO4 and 0.5 water, this works out 1:2. Methodically, divide by the sma number to work out the ratio as you often get numbers.


Original post
by tomsey11
Remember no mass can be lost or gained throughout the reaction. With this in mind we can find out the mass of the water on the right side of the equation:
4.92 - 2.40 = 2.52.

The molar mass of water (H2O) is 18 (1+1+16) so to find the number of moles of it you do the molar mass divided by the actual mass:

18.00 / 2.52 = 7.14

The value will never be perfect because of experimental error, so round it to 7. So there are 7 moles of water on the right side of the equation, so n = 7.

Hope that helps

Thanks I had a similar question and you really helped