Question

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

Answer 1

Answer:

65.2 g of  $Na_{2}SO_{3}$ and 9.3 g of $H_{2}O$ were formed

Explanation:

Balanced reaction:

$SO_{2}+2NaOH\rightarrow Na_{2}SO_{3}+H_{2}O$

Molar mass of $SO_{2}$ = 64 g

Molar mass of NaOH = 40 g

Molar mass of $Na_{2}SO_{3}$ = 126 g

Molar mass of $H_{2}O$ = 18 g

36.8 g of $SO_{2}$ = $\frac{36.8}{64}moles$ of $SO_{2}$ =  0.575 moles of $SO_{2}$

20.7 g of NaOH = $\frac{20.7}{40}moles$ of NaOH = 0.5175 moles of NaOH

According to balanced equation-

2 moles of NaOH react with 1 mol of $SO_{2}$ to produce 1 mol of $Na_{2}SO_{3}$ and 1 mol of $H_{2}O$ .

So, 0.5175 moles of NaOH react with $\frac{0.5175}{2}$ moles of $SO_{2}$ or 0.25875 moles of $SO_{2}$ to produce 0.5175 moles of $Na_{2}SO_{3}$ and 0.5175 moles of $H_{2}O$.

So, mass of $Na_{2}SO_{3}$ formed = ($0.5175\times 126$) g = 65.2 g

mass of $H_{2}O$ formed = ($0.5175\times 18$) g = 9.3 g