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you have 20 g of palladium and 18 g of magnesium sulphide. 2Pd + 3MgS a. What is the limiting reactant? 3Mg + Pd, S, b. Ho w much excess of the other reactant will you have? What substance (s) will be left after this reaction is finished?

you have 20 g of palladium and 18 g of magnesium sulphide. 2Pd + 3MgS a. What is the limiting reactant? 3Mg + Pd, S, b. Ho w much excess of the other reactant will you have? What substance (s) will be left after this reaction is finished?

Grade:11

1 Answers

Kapil Khare
80 Points
5 years ago
The balanced chemical reaction is
                    2Pd + 3MgS  \rightarrow 3Mg + Pd2S3
 
Number of moles of palladium = 20/(106) = 0.19moles
Number of moles of magnesium sulphide = 18/(56) = 0.32moles
 
1 mole of Palladium uses 1.5 moles of magnesium sulphide then,
Number of moles of magnesium sulphide used by given mass of palladium = 1.5*(0.19) = 0.285moles
\implies Palladium is the limiting reagent
 
Number of moles of magnesium sulphide left = 0.32 – 0.285 = 0.035moles
Mass of magnesium sulphide left = 0.035*(56) = 1.96grams
 
Magnesium, Palladium sulphide and magnesium sulphide will be left after the reaction is finished.

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