How many grams of Na2SiO3 is obtained when 3.25 g of silicon is added to sodium hydroxide : Si + 2NaOH + H2O -->  Na2SiO3 + H2

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justaguide | College Teacher | (Level 2) Distinguished Educator

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When silicon is added to a sodium hydroxide solution the reaction that takes place is : Si + 2NaOH + H2O -->  Na2SiO3 + 2H2

One mole of silicon reacts with 2 moles of sodium hydroxide to give one mole of Na2SiO3. The molar mass of silicon is 28.0855 g/mol. 3.25 g of silicon is equivalent to 3.25/28.0855 = 0.1157 mole of silicon. When 0.1157 mole of silicon takes part in the reaction 0.1157 moles of Na2SiO3 is formed. The molar mass of Na2SiO3 is 122.0634 g/mol. 0.1157 moles of Na2SiO3 is equivalent to 0.1157*122.0634 = 14.12 g of Na2SiO3.

When 3.25 g of silicon takes part in the given reaction, 14.12 g of Na2SiO3 is formed.

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