CH\(_3\) - C≡CH +  2Br\(_2\)  → CH\(_3\)-CBr\(_2\) -CH(Br\(_2\))

  1 mole  :  2 moles

Recall that n = \(\frac{m}{M}\) 

Molar mass of Br\(_2\) = 80 x 2 = 160 g/mol

 n = \(\frac{8}{160}\)

 n = 0.05 mole

If  1 mole of CH\(_3\) - C≡CH reacts with 2 moles of Br\(_2\) 

then X mole of CH\(_3\) - C≡CH will react with 0.05 mole of Br\(_2\) 

  X = \(\frac{0.05}{2}\)

  X = 0.025 mole of CH\(_3\) - C≡CH

Also, recall that n = \(\frac{m}{M}\) 

Molar mass of CH\(_3\) - C≡CH = C3H4 = (12 X 3) + (1 X 4) = 40 g/mol

n = \(\frac{m}{M}\)

n = 0.025 x 40 = 1.0 g