Aqueous sodium propanoate ( $CH X_{3} CH X_{2} COONa$ , molar mass $96 g mo l^{- 1}$ ) 19.2 g is electrolysed in a Kolbe synthesis. The alkane formed at the anode is obtained in 75\% yield based on the stoichiometry of coupling. Molar mass of this alkane is $58 g mo l^{- 1}$ . Find the mass of the alkane in grams (two decimals).

Numerical answer type, answer in g, tolerance 0.05.

Answer: 4.35 g

Two propanoate ions give one molecule of

CH X_{3} CH X_{2} - CH X_{2} CH X_{3}

(butane). Moles of salt

= 19.2/96 = 0.20

, so theoretical butane

= 0.100

mol. At 75\% yield:

0.100 \times 0.75 = 0.0750

mol, mass

= 0.0750 \times 58 = 4.35