How many dm³ of hydrogen, measured at s. t.p.,would be needed to reduce 47.7g of copper(II) oxide to copper? A. 4.48
B. 6.72
C. 10.82
D. 13.44

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13.44 dm³ of hydrogen is needed for the reaction.

(D) is correct option

Explanation:

Given that,

Mass of  copper(II) oxide = 47.7 g

We know that,

Molar mass of CuO = 63.5 + 16 = 79.5 g/mol

We need to calculate the number of moles in 47.7g of copper(II) oxide,

Using formula of moles

Put the value into the formula

Now, we write the balanced equation for the reaction

Here, one mole of CuO reacted with one mole of hydrogen to produce one mole of Cu and one mole of water.

We need to calculate the number of moles of hydrogen needed to react  with 0.6 mole of copper(II) oxide

Using the balanced equation,

1 mole of CuO reacted with 1 mole of Hydrogen.

Therefore, 0.6 mole of CuO will also react with 0.6 mole of hydrogen.

We need to calculate the volume of  occupied by 0.6 mole of H2 at STP

Using formula of volume

Where, n = mole of hydrogen

V = occupied volume

V'= volume at STP

Put the value into the formula

Hence, 13.44 dm³ of hydrogen is needed for the reaction.