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Limiting Stoichiometry--PLEASE HELP!!!?
Hi guys, I need help with limiting stoichiometry. Here are the questions:
1) Zinc metal of 2.00 grams is placed in a solution containing 2.50 grams of silver nitrate.
a) find the limiting reactant
b)how many grams of the silver containing product will form?
c)how many grams of the excess reactant will be left over at the end of the reaction?
d)if only 0.45 grams of the silver containing product form, calculate the percent yield.
Please put the letter of the question you answered, that'd be great. Also, can someone please explain to me how you get b and c, 'cause that's what lost me. Thanks to all who help me!!!
1 Answer
- Anonymous1 decade agoFavorite Answer
Zn + 2AgNO3 >> Zn(NO3)2 + 2Ag
a) 2.00 g / 65.37 g/mol= 0.0306 moles Zn
2.50 g/ 231.87 g/mol = 0.0108 moles Ag(NO3)2
Ag(NO3)2 limiting reactant
b) moles of silver = 0.0108
g of silver = 0.0108 mole (107.87 g/mol) = 1.16 g
c) Since the ratio between Zn and Ag(NO3)2 is 1 : 2
1 : 2 = x : 0.0108
x = 0.00540 moles Zn needed for the reaction
0.0306 - 0.00540 = 0.0252 moles Zn in excess
0.0252 mole ( 107.87 g/mol)= 2.72 g Zn in excess
d) 0.45 : 1.16 = x : 100
x = 38.8 %