Sulfur dioxide gas reacts with sodium hydroxide to form sodium sulfite and water. the unbalanced chemical equation for this reaction is given below: so2(g) + naoh(s) → na2so3(s) + h2o(l)assuming that you start with 36.8 g of sulfur dioxide and 20.7 g of sodium hydroxide and assuming that the reaction goes to completion, determine the mass of each product formed. g na2so3g h2o

65.2 g of   and 9.3 g of were formed

Explanation:

Balanced reaction:

Molar mass of = 64 g

Molar mass of NaOH = 40 g

Molar mass of = 126 g

Molar mass of = 18 g

36.8 g of = of =  0.575 moles of

20.7 g of NaOH = of NaOH = 0.5175 moles of NaOH

According to balanced equation-

2 moles of NaOH react with 1 mol of to produce 1 mol of and 1 mol of .

So, 0.5175 moles of NaOH react with moles of or 0.25875 moles of to produce 0.5175 moles of and 0.5175 moles of .

So, mass of formed = () g = 65.2 g

mass of formed = () g = 9.3 g

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explanation: