Question

30 ml of. 150 m cacl2 is added to a 15 ml of. 100 m agno3

Answer 1

The mass of the formed precipitate of AgCl in the reaction is 1.29 grams.

How do we find moles from molarity?

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.

To know more about moles & molarity, visit the below link:
brainly.com/question/24322641

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