Answer to Question #180684 in General Chemistry for Enci Peng

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

From the equation

2 moles of Au(NO₃)_3(aq) = 3 moles of Fe

= 6.5 ×10²³/6.022 × 10²³ = 1.0794 moles

= 1.0794 × 2/3 = 0.7196

Molar Mass of gold nitrate = 382.98

= 0.7196 × 382.98

= 275.59g

= 0.2756 L