Calculate the mas of copper sulphate when 2.5g of copper(II) oxide reacts with excess sulphuric acid. The equation for the reaction is:-
- CuO + H2SO4 --> CuSO4 + H2O
Ar:
Cu= 63.5
O= 16
S= 32
Can you please help me solve this problem, i heard there was two ways of solving it, one is by '' using relative formula mases '' and the other is to use " moles ". If you can put down the two methods down that will help allot, if you can just put down the easier method... and put down how to check your answer...,
Thanks
- CuO + H2SO4 --> CuSO4 + H2O
Ar:
Cu= 63.5
O= 16
S= 32
Can you please help me solve this problem, i heard there was two ways of solving it, one is by '' using relative formula mases '' and the other is to use " moles ". If you can put down the two methods down that will help allot, if you can just put down the easier method... and put down how to check your answer...,
Thanks
-
79.5 g of CuO forms = 159.5 g of CuSO4
1 g of CuO forms = 159 . 5 / 79.5 g of CuSO4
2.5 g of CuO forms = ( 159 . 5 / 79 . 5 ) x 2.5
= 5.015 g of CuSO4
1 g of CuO forms = 159 . 5 / 79.5 g of CuSO4
2.5 g of CuO forms = ( 159 . 5 / 79 . 5 ) x 2.5
= 5.015 g of CuSO4