Asked by Denmark
                A student was asked to make some copper(II) sulfate-5-5 (CuSO4.5(H2O) BY reacting copper(II) oxide (CuO) with sulfuric acid.
a) Calculate the molar mass of the copper (II) sulfate-5-water
which is (64+32+64) + (10+80)=250 mol
b)Calculate the amount (in mol) of copper (II) sulfate-5-water in a 10,0g sample?
C) Calculate the mass of copper (II) oxide needed to make the 10,0g sample?
Can anyone help me with question b,c??
            
        a) Calculate the molar mass of the copper (II) sulfate-5-water
which is (64+32+64) + (10+80)=250 mol
b)Calculate the amount (in mol) of copper (II) sulfate-5-water in a 10,0g sample?
C) Calculate the mass of copper (II) oxide needed to make the 10,0g sample?
Can anyone help me with question b,c??
Answers
                    Answered by
            Anonymous
            
    Amount of moles= mass/molar mass
No. of moles= 10.0g/250 g
No. of moles= 0.04 moles
c) CuO + H2SO4 ----> CuSO4 + H2O
mol 1 1 1 1
1 mole of CuO = 1 mole of CuSO4.5H2O
CuO= 64g + 16g= 80g
80 g of CuO = 250 g of CuSO4
x = 10g of CuSO4
10g x 80g/250g
800g/250
=3.2 grams of CuO
    
No. of moles= 10.0g/250 g
No. of moles= 0.04 moles
c) CuO + H2SO4 ----> CuSO4 + H2O
mol 1 1 1 1
1 mole of CuO = 1 mole of CuSO4.5H2O
CuO= 64g + 16g= 80g
80 g of CuO = 250 g of CuSO4
x = 10g of CuSO4
10g x 80g/250g
800g/250
=3.2 grams of CuO
                    Answered by
            Denmark
            
    thanks you very much.
    
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