what mass of silver chloride can be produced from 1.83 L of a 0.148 M solution o

Question

what mass of silver chloride can be produced from 1.83 L of a 0.148 M solution of silver nitrate? and the reaction described in Part A required 3.75 L of calcium chloride. what is the concentration of this calcium chloride solution? When solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution according to the equation 2AgNO3 (aq) + CaCl2 (aq)--2AgCl (s) + Ca(NO3 ) 2 (aq) Part A What mass of silver chloride can be produced from 1.83 L of a 0.148 M solution of silver nitrate? Express your answer with the appropriate units LA ? mass of AgcI-Value Units mass of AgCl = 1 Value Units Submit Hints My Answers Give Up Review Part Part B The reaction described in Part A required 3.75 L of calcium chloride. What is the concentration of this calcium chloride solution? Express your answer with the appropriate units Value Units Submit Hints My Answers Give Up Review Part

Explanation / Answer

Part A:
Number of moles of AgNO3 = M*V = 0.148 M * 1.83 L = 0.27084 mol

2 mol of AgNO3 gives 2 moles of AgCl

So,number of moles of AgCl produced = 0.27084 mol

Molar mass of AgCl = 143.32 g/mol

Mass of AgCl produced = number of moles * molar mass
                                                = 0.27084 mol*143.32 g/mol
                                                 =38.82 g
Answer: 38.82 g

Part B :
Number of moles of AgNO3 = 0.27084 mol
2 mol of AgNO3 reacts with 1 moles of CaCl2
so number of moles of CaCl2 required = 0.27084/2 = 0.13542 mol

use:
number of moles = M*V
0.13542 = M*3.75
M=0.036 M

Answer: 0.036 M

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