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Calculate the volume in mL of 1.50 M BaCl2 required to produce 57.2 g. of AgCl.

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Calculate the volume in mL of 1.50 M BaCl2 required to produce 57.2 g. of AgCl acording to the following chemical equation:

AgNo^3(aq) + BaCl^2(aq) - AgCl(s) + Ba(No^3)^2(aq)

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Solution Summary

According to the chemical equation 1 mole of BaCl2 produces 1 mole of AgCl. Therefore, knowing the molecular weights we can determine how many moles of BaCl2 are needed to produce 57.2g of AgCl. Similarly, knowing 1.5M BaCl2 means 1.5 moles of BaCl2/lit solution we can determine how much volume of solution contains the required moles of BaCl2.

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Because, according to the chemical equation, 1 mole of BaCl2 produces 1 mole of AgCl:

molecular wt. of AgCl = 108+35.5 = 143.5
molecular wt. of BaCl2 = 137+35.5*2 = ...

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