Question

The number of moles of CaCl2 needed to react with excess of AgNO3 to produce 4.31 gram of AgCl​

Answer 1

Answer:

2 AGNO3 + CACl2 = 2 AGCl + CA(NO3)2

Molar mass of AgCl= 143.32

Number of moles of 4.31g of AgCl= 0.03mol

From the reaction, 1mole of CaCl2 produced 2mole of AgCl

Hence, x mole of CaCl2 produced 0.03mole of AgCl.

x=0.015mol

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Answer 2

Answer:

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