First slide

Stoichiometry and Stoichiometric Calculations

Question

60 grams of sodium hydroxide is required to completely neutralize 49 grams of sulphuric acid, percentage purity of sodium hydroxide sample is

Moderate

A

66.6%
B

33.3%
C

54%
D

90%

Solution

$2 NaOH + H_{2} {SO}_{4} \to {Na}_{2} {SO}_{4} + 2 H_{2} O; (2 \times 40) g 98 g 80 grams of pure NaOH is required to neutralize 98 grams of H_{2} {SO}_{4}; 40 grams of pure NaOH is required to neutralize 49 grams of H_{2} {SO}_{4}; But in the given question, 60 grams of NaOH was required; Percentage purity of NaOH = \frac{40}{60} \times 100 = 66.6 %$

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