How many grams of Na3AsO3 is produced from the reaction of 6.90 moles of NaOH?

Question

Chemistry

Hadassah Fuller

14 November, 13:55

How many grams of Na3AsO3 is produced from the reaction of 6.90 moles of NaOH?

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Answers (1)

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Jayden Gay · Answer 1

6.90 mol x grams

2As + 6NaOH = 2Na3AsO3 + 3H2

6 mol 2 mol

192 g/mol

6.90 mol NaOH x 2 mol Na3AsO3/6 mol NaOH= = 2.3 mol Na3AsO3

2.3 mol Na3AsO3 x 192 g/mol = 442 g Na3AsO3