Calculate the mass of copper(II) sulfate formed when $2.5 g$ of copper (II) oxides react with excess sulfuric acid. The equation for the reaction is:
$C u O + H_{2} {S O}_{4} \to {C u S O}_{4} + H_{2} O$
(Atomic number:Cu=63; O=16; S-32)

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Solution

Solution:-
$C u O + H_{2} S O_{4} ⟶ C u S O_{4} + H_{2} O$
Mol. wt. of $C u O = 79 g$
Mol. wt. of $C u S O_{4} = 159 g$
Mass of copper(II) sulfate formed when 79 gm of copper(II) oxide react $= 159 g$
Mass of copper(II) sulfate formed when 2.5 gm of copper(II) oxide react $= \frac{159}{79} \times 2.5 = 5.031 g$
Hence, 5.031 gm of $C u S O_{4}$ formed when 2.5 gm of $C u O$ reacts with excess $H_{2} S O_{4}$ .

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Similar questions

Calculate the mass of copper(II) oxide required when reacts with excess sulfuric acid to form 1 mol of $C u S O_{4}$ . The equation for the reaction is:

$C u O + H_{2} S O_{4} \to C u S O_{4} + H_{2} O$

[Given $: C u = 63.5; O = 16; S = 32$ ]

Q. The product formed along with copper sulfate on reaction of copper with sulfuric acid is:

From the equation

C + 2 H_{2} {S O}_{4} \to {C O}_{2} + {2 H}_{2} O + 2 {S O}_{2}

. Calculate the mass of the carbon oxidized by 49g of sulphuric acid. (Atomic mass of C=12, H=1, S=32, O=16)

Q. Obtaining copper sulfate from copper oxides by using sulfuric acid is a:

Q. Calculate the percentage loss of mass of hydrated copper[II] sulphate

[C u S O_{4} .5 H_{2} O]

when it is completely dehydrated.

C u S O_{4} .5 H_{2} O \to C u S O_{4} + 5 H_{2} O

[Atomic weights are

C u = 64, S = 32, O = 16, H = 1

]

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Calculate the mass of copper(II) sulfate formed when 2.5g of copper (II) oxides react with excess sulfuric acid. The equation for the reaction is:CuO+H2SO4→CuSO4+H2O(Atomic number:Cu=63; O=16; S-32)

Calculate the mass of copper(II) sulfate formed when $2.5 g$ of copper (II) oxides react with excess sulfuric acid. The equation for the reaction is:
$C u O + H_{2} {S O}_{4} \to {C u S O}_{4} + H_{2} O$
(Atomic number:Cu=63; O=16; S-32)