The mass of the formed precipitate of AgCl in the reaction is 1.29 grams.
<h3>How do we find moles from molarity?</h3>
Moles (n) of any substance from molarity (M) will be calculated by using the below equation:
M = n/V, where
V = volume in L
Given chemical reaction is:
2AgNO₃(aq) + CaCl₂(aq) → 2AgCl(s) + Ca(NO₃)₂(aq)
- Moles of CaCl₂ = 0.150M × 0.03L = 0.0045 moles
- Moles of AgNO₃ = 0.100M × 0.015L = 0.0015 moles
From the stoichiometry of the reaction, mole ratio of AgNO₃ to CaCl₂ is 2:1.
0.0015 moles of AgNO₃ = reacts with 1/2×0.0045 = 0.00075 moles of CaCl₂
Here CaCl₂ is the limiting reagent, and formation of precipitate depends on this only.
Again from the stoichiometry of the reaction:
0.0045 moles of CaCl₂ = produces 2(0.0045) = 0.009 moles of AgCl
Mass of 0.009 moles AgCl will be calculated as:
n = W/M, where
- W = required mass
- M = molar mass = 143.45 g/mol
W = (0.009)(143.45) = 1.29g
Hence required mass of precipitate is 1.29 grams.
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