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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?

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  • 2 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

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  • 2 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 =

    9.10329gS needed

    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.

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  • 2 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.

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