Answer to Question #181004 in General Chemistry for Sophia Beer

Question

Throwing some scrap iron in a gold nitrate solution causes the gold metal to precipitate. How many liters of 1 M gold nitrate solution would react with 6.5x10²³ particles of iron metal?

3 Fe_(s) + 2 Au(NO₃)_3(aq) --> 3 Fe(NO₃)_2(aq) + 2 Au_(s)

Answer 1

no. of mole of Fe = $\frac{6.5×10^{23}}{6.022×10^{33}}$ = 1.08 mol

for 3 mol of Fe = 2 mol of Au(NO₃)₃

1.08 mol of Fe = $\frac{2}{3} × 1.08$

= 0.72 mol

1 M = (1 mol / 1 L)

1 mol = 1 L

0.72 mol = 0.72 L