Answer:
454 grams AgCl2 (3 sig figs)
Step-by-step explanation:
2AgNO3(aq) + MgCl2(aq) → 2AgCl(s) + Mg(NO3)2(aq)
the equation appears balanced. It tells us we need 2 moles of AgNO3 for every 1 mole of MgCl2 to form 2 moles of AgCl and 1 mole of Mg(NO3)2.
What mass of silver chloride can be produced from 1.22 L of a 0.208 M solution of silver nitrate? Express your answer with the appropriate units.
Lets determine how many moles of AgNO3 are in 1.22L of 0.208M solution of AgNO3. Remember that M stands for (moles/liter). Also remember that the moles of a substance can be found by multiply the volume (V) times the concentration (M). Change M to moles/liter to insure the units cancel correctly:
moles AgNO3 = (1.22liters)*(2.08 moles/liter).
moles AgNO3 = 2.538 moles AgNO3
The balanced equation tells us that 2 moles of AgNO3 will produce 2 moles of AgCl, assuming there is sufficient MgCl2.
That is a 1:1 molar ratio. What we put in as moles of AgNO3 should produce the same moles of AgCl2. We would expect to find 2.538 moles of AgCl2.
To find that amount in grams, convert the moles AgCl2 to grams AgCl2 using the molar mass of AgCl2, 178.8 grams/mole.
(2.538 moles of AgCl2)*(178.8 grams/mole) = 454 grams AgCl2 (3 sig figs)