I do these limiting reagent (LR) problems the long way. There are shorter ways to do them.
2Na + Br2 --> 2NaBr
mols Na = grams/atomic mass = ?
mols Br2 = grams/molar mass
Using the coefficients in the balanced chemical equation, convert mols Na to mols NaBr.
Do the same for mols Br2 to mols NaBr.
It is likely these two values will not agree which means one of them is not right; the correct value in LR problems is ALWAYS the smaller number and the reagent producing that number is the LR. That goves up a.
c you get by a x molar mass NaBr = g NaBr
b. Using the coefficients in the balanced equation, convert mols of the LR to moles of the OR (other reagent). That gives you the mols of the other reagent used. Then excess mols = initial mols - mols used = final mols.
Convert that to grams by g = mols x atomic mass.
If 80.0 g of bromine is added to 40.0 g of sodium,
a) what is the limiting reactant?
b) what is the excess mass of the reactant?
c) what is the mass of product?
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