In a lab experiment, 6.9 g of Mg reacts with sulfur to form 16.0 g of magnesium sulfide. How much magnesium sulfide would be formed if 6.9 g of Mg were reacted with 17.9 g of S?
- Roger the MoleLv 710 months ago
Mg + S → MgS
(6.9 g Mg) / (24.30506 g Mg/mol) = 0.28389 mol Mg
(17.9 g S) / (32.0655 g S/mol) = 0.558232 mol S
0.28389 mole of Mg would react completely with 0.28389 mole of S, but there is more S present than that, so S is in excess and Mg is the limiting reactant.
(0.28389 mol Mg) x (1 mol MgS / 1 mol Mg) x (56.370 g MgS/mol) = 16 g MgS
- Anonymous10 months ago
The chemical reaction equation is:
Mg + S --> MgS so the mole ratios for everything is 1
6.9g * 1mol/24.305g = 0.2838mol Mg
17.9g * 1mol/32.066g = 0.55822mol S
To find the limiting reactant:
6.9gMg/1 * 1molMg/24.305gMg * 1molS/1molMg * 32.066gS/1molS =
You have 17.9gS so Mg is the limited reactant.
You use the limited reactant to do dimensional analysis:
6.9gMg/1 * 1molMg/24.305gMg * 1molMgS/1molMg * 56.38gMgs/1molMgs =
16.0058g MgS formed
You don't even need the first part of the question.
- az_lenderLv 710 months ago
MgS has equal molar amounts of Mg and S.
The atomic mass of Mg is 24.305 g/mol;
the atomic mass of S is 32.065 g/mol.
The 6.9 g of Mg is a little more than a quarter of a mole. If sulfur were supplied in "excess," the amount of MgS expected would be
(6.9 g)*(24.3 + 32.1)/(24.3) = 16.0 grams.
This is the MOST MgS you can get out of 6.9 g of magnesium, so the answer to the question is still 16.0 g.