Answer:
23.2 g of Al will be left over when the reaction is complete
Explanation:
2Al + 3S → Al₂S₃
1 mol of Al = 26.98 g
1 mol of S = 32.06 g
Mole = Mass / Molar mass
63.8 g/ 26.98 g/m = 2.36 mole of Al
72.3 g / 32.06 g/m = 2.25 mole of S
2 mole of Aluminun react with 3 mole of sulfur
2.36 mole of Al react with (2.36 .3)/2 = 3.54 m of S
As I have 2.25 mole of S, and I need 3.54 S, is my limiting reagent so the limiting in excess is the Al.
3 mole of S react with 2 mole of Al
2.25 mole of S react with (2.25 m . 2)/3 = 1.50 mole
I need 1.50 mole of Al and I have 2.36, that's why the Al is in excess.
2.36 mole of Al - 1.50 mole of Al = 0.86 mole
This is the quantity of Al without reaction.
Molar mass . mole = Mass → 26.98 g/m . 0.86 m = 23.2 g
Only answers C and D would make sense in this question. Hope I helped please give brainiest!
It is fresher because of new transportation techniques and refrigerated trailers.
The error is in the formula is bracket so the correct is Mg(OH)₂
The incorrect or error in the formula is bracket that would be that the compound contain only 1Mg atom and 1 oxygen atom and 2 hydrogen atom but is wrong when you write it in an electron dot diagram you will see that it is wrong there be unshared paired of electron and the correct formula is Mg(OH)₂
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