Answer:
r = k [ B ]
Explanation:
⇒ r = k [ A ]/2 = k [ B ]
Mass of Na2SO4= 514.18 grams
<h3>Further explanation</h3>
Given
423.67 g of NaCl
Required
mass of Na2SO4
Solution
Reaction
2NaCl + H2SO4 → Na2SO4 + 2HCl
mol NaCl :
= 423.67 g : 58.5 g/mol
= 7.24
From the equation, mol Na2SO4 :
= 1/2 x mol NaCl
= 1/2 x 7.24
= 3.62
Mass Na2SO4 :
= 3.62 mol x 142,04 g/mol
= 514.18 grams
FeBr₃ ⇒ limiting reactant
mol NaBr = 1.428
<h3>Further explanation</h3>
Reaction
2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr
Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)
211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :

So FeBr₃ as a limiting reactant(smaller ratio)
mol NaBr based on limiting reactant (FeBr₃) :
