[abhishek007]

a

[abhishek007]

a

[Dan]

what i did was....

a. The stream of hydrogen was maintained as the hydrogen would still react until the apparatus completely cooled off.

No. You have reduced the copper oxide to copper, once the reaction is complete heating is stopped. Copper(0) will not react with hydrogen. The hydrogen gas flow is maintained to prevent oxygen getting into the tube, which would react with the hot copper and regenerate some of the copper oxide - undoing your hard work.

b.
Mass copper oxide at beginning=7.95g
afterwords: 6.35g///

So how many moles of Cu are present at the end?
How many moles were present at the start?
Some mass was lost - this is the mass of the oxygen, and how many moles of it were lost?
So how many moles of it were present in the copper oxide?
The ratio of your moles of Cu to O gives the empirical formula.

So now write an equation.

d. The hydrogen burning will have a green color.

Also water droplets might appear.

Water droplets sound reasonable (consider this when you work out your equation for c).
As far as I know, the flame from burning hydrogen is pretty much invisible. Anyway, the flame is outside the tube.
Think about the starting material compared to the compound. Will you notice any physical changes - like colour change, phase change (melting etc.), or anything else?

[abhishek007]

a

[Dan]

   
B.

7.95g / 63.54g (molar mass) = 0.125118036 mol copper oxide  (at the start)

This bit does not make sense; at the start you have 7.95 g of a copper oxide, you cannot determine how many moles of it you have by dividing by the relative atomic mass of Cu!
Anyway, what you have after that statement is fine - it is CuO, and all the working looks fine. Your equation looks good too (apart from the capital U typo).

For D., I agree on the colour change observation but you should really specify the colour change you would expect.

[abhishek007]

thanks so much for your help Dan