**If a solution contains 4.9g of tetraoxosulphate (VI) acid, calculate the amount ...**

### Question

If a solution contains 4.9g of tetraoxosulphate (VI) acid, calculate the amount of copper (II) oxide that will react with it

[Cu = 64, O = 16, S = 32, H = 1]

### Options

A) 0.8g

B) 4.0g

C) 40.0g

D) 80.0g

Related Lesson: Producing Sulfuric Acid | Chemistry and the Real World

The correct answer is B.

### Explanation:

Equation of the reaction

H

Relative molecular mass of H

Relative molecular mass of CuO = 64 + 16 =80

From the above equation, 98

∴ 4.9g of H

= (4.9 * 80)/98 of CuO = 392/98 = 4.0g

H

_{2}SO_{4(g)}+ CuO_{(s)}→ Cu_{4(aq) + H}_{2}ORelative molecular mass of H

_{2}SO_{4}= 1 + 2 + 32 + 16 + 4 = 98Relative molecular mass of CuO = 64 + 16 =80

From the above equation, 98

_{s}of H_{2}SO_{4}reacts with 80_{s}of CuO.∴ 4.9g of H

_{2}SO_{4}will react with= (4.9 * 80)/98 of CuO = 392/98 = 4.0g

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Equation of the reaction

H

_{2}SO_{4(g)}+ CuO_{(s)}→ Cu_{4(aq) + H}_{2}ORelative molecular mass of H

_{2}SO_{4}= 1 + 2 + 32 + 16 + 4 = 98Relative molecular mass of CuO = 64 + 16 =80

From the above equation, 98

_{s}of H_{2}SO_{4}reacts with 80_{s}of CuO.∴ 4.9g of H

_{2}SO_{4}will react with= (4.9 * 80)/98 of CuO = 392/98 = 4.0g