What mass of metal is formed by the interaction of iron with silver nitrate (1) in a solution
August 18, 2021 | education
| What mass of metal is formed by the interaction of iron with silver nitrate (1) in a solution weighing 320 g with a mass fraction of salt of 0.2?
1. Iron is a more active metal in comparison with silver, therefore it displaces it from the salt solution:
Fe + 3AgNO3 = Fe (NO3) 3 + 3Ag;
2. Calculate the mass of silver nitrate:
m (AgNO3) = w (AgNO3) * m (solution) = 0.2 * 320 = 64 g;
3. Let’s calculate the chemical amount of silver nitrate:
n (AgNO3) = m (AgNO3): M (AgNO3);
M (AgNO3) = 108 + 14 + 3 * 16 = 170 g / mol;
n (AgNO3) = 64: 170 = 0.3765 mol;
4. Let’s set the amount of the formed metal:
n (Ag) = n (AgNO3) = 0.3765 mol;
5. Find the mass of silver:
m (Ag) = n (Ag) * M (Ag) = 0.3765 * 108 = 40.66 g.
Answer: 40.66 g.
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