spastik08
contestada

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.

How many moles of silver would precipitate out when 36 x 1023 atoms of zinc are combined in silver nitrate according to the following unbalanced equation Balanc class=

Respuesta :

maacastrobr

Answer:

D: 0.6 moles

Explanation:

First we balance the equation:

  • Zn + AgNO₃ → Zn(NO₃)₂ + Ag

There are two NO₃ species on the right side, so we put a two before  AgNO₃:

  • Zn + 2AgNO₃ → Zn(NO₃)₂ + Ag

Now we put a two before Ag on the right, to balance the Ag atoms:

  • Zn + 2AgNO₃ → Zn(NO₃)₂ + 2Ag

The 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 moles

Then 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 Ag

Thus the answer is option D.

Otras preguntas