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.

D: 0.6 moles

Explanation:

First we balance the equation:

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

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

The equation is now balanced.

Now we convert 3.6x10²³ atoms of zinc to moles, using Avogadro's number:

Then we convert 0.6 moles of zinc into moles of silver, using the stoichiometric coefficients of the balanced equation:

Thus the answer is option D.

how to make a simple air condition

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g