How many moles of silver would precipitate out when 3.6 x 10^23 atoms of zinc are combined in silver nitrate according to the following unbalanced equation? (Balance equation first!)
Zn + AgNO3 → Zn(NO3)2 +Ag
A: 1.67 moles
B: 1.2 moles
C: 3.34 moles
D: 0.6 moles
Can someone please help me with this I don’t get it at all.
Answers
Answer:
D: 0.6 moles
Explanation:
First we balance the equation:
Zn + AgNO₃ → Zn(NO₃)₂ + AgThere are two NO₃ species on the right side, so we put a two before AgNO₃:
Zn + 2AgNO₃ → Zn(NO₃)₂ + AgNow we put a two before Ag on the right, to balance the Ag atoms:
Zn + 2AgNO₃ → Zn(NO₃)₂ + 2AgThe equation is now balanced.
Now we convert 3.6x10²³ atoms of zinc to moles, using Avogadro's number:
3.6x10²³ atoms ÷ 6.023x10²³ atoms/mol = 0.6 molesThen we convert 0.6 moles of zinc into moles of silver, using the stoichiometric coefficients of the balanced equation:
0.6 mol Zn * [tex]\frac{1molAg}{1molZn}[/tex] = 0.6 mol AgThus the answer is option D.
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Answer:
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Credits go to the person above me.
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EDGE 2021
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