Correct Answer: Option B

Explanation

Equation of the reaction
H\(_2\)SO\(_4\)(g) + CuO(s) → CuSO\(_4\)(aq) + H\(_2\)O
Relative molecular mass of H\(_2\)SO\(_4\) = 1 + 2 + 32 + 16 + 4 = 98
Relative molecular mass of CuO = 64 + 16 =80
From the above equation, 98g of H\(_2\)SO\(_4\) reacts with 80g of CuO.
∴ 4.9g of H\(_2\)SO\(_4\) will react with Xg of CuO
= (4.9 * 80)/98 of CuO = 392/98 = 4.0g

Alternatively,

H\(_2\)SO\(_4\)(g) + CuO(s) → CuSO\(_4\)(aq) + H\(_2\)O

   1           :        1

Since the combining ratio of  H\(_2\)SO\(_4\) and CuO  is in ratio 1:1, the number of moles will also be equal.

Hence, we can first calculate the amount of H\(_2\)SO\(_4\), since we were given the mass of H\(_2\)SO\(_4\) that reacted as 4g and the Molecular mass is 98g/mol

Amount of moles of H\(_2\)SO\(_4\) = \(\frac{mass}{Molecular mass}\)

                                                         =  \(\frac{4.9}{98}\)  = 0.05 mole

Amount of moles of CuO =  \(\frac{mass}{Molecular mass}\)

                                      0.05  =  \(\frac{mass}{64 + 16}\)

Mass = 0.05 x 80 = 4.0g

There is an explanation video available below.

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