A mixture of $Na X_{2} CO X_{3}$ and $NaHCO X_{3}$ contains $40.0%$ sodium by mass (Na = 23, C = 12, O = 16, H = 1). What is the mass percentage of $Na X_{2} CO X_{3}$ in the mixture? Give one decimal place.

Numerical answer type, answer in %, tolerance 0.1.

Answer: 78.8 %

Mass fraction of Na is

46/106 = 0.4340

Na X_{2} CO X_{3}

and

23/84 = 0.2738

NaHCO X_{3}

. With

w

the fraction of

Na X_{2} CO X_{3}

0.4340 w + 0.2738 (1 - w) = 0.400

, so

w = (0.400 - 0.2738) / (0.4340 - 0.2738) = 0.7879

, i.e.

78.8%