Throwing some scrap iron in a gold nitrate solution causes the gold metal to precipitate. How many liters of 0.5 M gold nitrate solution would react with 2.8x1023 particles of iron metal?
Molar Mass of gold nitrate = 382.98g/mol
0.5 gold nitrate = 0.5×382.98 = 191.49g
2.8 × 10^23/6.022 × 10^23
= 0.465 × 191.49 = 89.0355g
= 0.0890355 Litres of 0.5M gold nitrate
Comments
Leave a comment