14.46g of sodium reacts with 10.00g of iron (III) sulfide. How many grams of excess reactant remains? 6 Na + Fe2S3 ->−> 3 Na2S + 2 Fe Chemical Molar mass(g/mol) Na 22.99 Fe2S3 207.91 Na2S 78.05 Fe 55.85
14.46g of sodium reacts with 10.00g of iron (III) sulfide. How many grams of excess reactant remains?
6 Na + Fe2S3 ->−> 3 Na2S + 2 Fe
| Chemical | Molar mass(g/mol) |
| Na | 22.99 |
| Fe2S3 | 207.91 |
| Na2S | 78.05 |
| Fe | 55.85 |
Given, mass of sodium, Na = 14.46 g
and, mass of iron (III) sulfide , Fe2S3 = 10.00 g
Now, moles of each reactant can be calculated as,
moles of Na = = = 0.629 mol
and, moles of Fe2S3 = = = 0.048 mol
The given reaction is,
6 Na + Fe2S3 --−> 3 Na2S + 2 Fe
Clearly, from the above reaction, we have,
1 mol of Fe2S3 requires = 6 mol of Na
so, 0.048 mol of Fe2S3 requires = 6 x 0.048 mol of Na
= 0.288 mol of Na
But, moles of Na reacted is 0.629 mol i.e. it is in excess.
So, the limiting reagent is Fe2S3 .
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