Sodium chlorite is an extremely powerful bleaching agent and numerous industrial

Question

Sodium chlorite is an extremely powerful bleaching agent and numerous industrial uses. Sodium chlorite is prepared by the reaction of sodium and calcium hydroxides with carbon and chlorine dioxide yielding sodium chlorite and calcium carbonate. First, write and balance the equation. Then calculate the percent yield of sodium chlorite if 1.000 kg of sodium hydroxide, 2.000 kg of calcium hydroxide. 0.500 kg of graphite, and 1.000 kg of chlorine dioxide are reacted yielding .750 kg of sodium chlorite.

Explanation / Answer

Balanced chemical equation for the reaction as follows :

Ca(OH)2 + NaOH + ClO2 + C = NaClO2 + CaCO3 + H2O

1000 gm NaOH × (1 mol NaOH / 39.997 gmNaOH) × (1 mol NaClO2 / 1 mol NaOH) = 25.00 mol of NaClO2

2000 gm Ca(OH)2 × (1 mol Ca(OH)2 / 74.09 gm Ca(OH)2) × (1 mol NaClO2 / 1 mol Ca(OH)2) = 26.99 mol of NaClO2

500 gm graphite × (1 mol graphite / 12.00 gm graphite) × (1 mol NaClO2 / 1 mol graphite) = 41.66 mol of NaClO2

1000 gm ClO2 × (1 mol ClO2 / 67.45 gm ClO2) × (1 mol NaClO2 / 1 mol ClO2) = 14.82 mol of NaClO2

Use the smallest number of moles of the product (NaClO2) from step 1 to calculate the theoretical yield of product (NaClO2).

theoretical yield :

= (14.82 mol of NaClO2) × (90.44 g/mol NaClO2 / 1 mol NaClO2 )

= 1340.8 gm of NaClO2

Practical yield :

0.750 kg means 750 gm of NaClO2 isolated from reaction

= (750 gm of NaClO2 / 1340.8 gm of NaCl2 theoretical) × 100 %

= 55.97 % percent yield of sodium chlorite

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