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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)
- FeBr₃
211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :
- Na₂S
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) :
Step 1 - Discovering the ionic formula of Chromium (III) Carbonate
Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):
Step 2 - Finding the molar mass of the substance
To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.
The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:
The molar mass will be thus:
Step 3 - Finding the percent composition of carbon
As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:
The percent composition of Carbon is thus 12.7 %.