Moles of ca3(po4)2 = 23.7 / 310.17 = 0.076
moles of (PO4)3- = 0.076 x 2 = 0.152
now, no. of ions = 0.152 x 6.022 x 10^{23}
= 9.2 x 10^{22}
Answer:
Look for extra things to do, small details, until you find a big enough one to go off that to continue
Explanation:
n/a
The balanced chemical reaction:
<span>Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
</span>
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.
9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3
The limiting reactant is AgNO3.
0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2
0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess
<span>0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess</span>
