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What mass of AgCl is produced when 53.42g of AgNO3 reacts with 14.19g of NaCl?1AgNO3+1NaCl=1AgCl+1NaNO3

Sagot :

Answer

the limiting reagent is NaCl

The mass of AgCl to be produced = 34.8 g

Explanation

Given:

Balanced chemical equation: 1AgNO3+1NaCl=1AgCl+1NaNO3

mass of AgCl = ?

AgNO3 = 53.42 g

NaCl = 14.19 g

Solution

Step 1: Calculate the number of moles of reactants, to find the limiting reactant/reagent.

For AgNO3:

[tex]53.42g\text{ AgNO}_3\text{ x }\frac{1\text{ mol AgNO}_3}{169,87g}\text{ x }\frac{1\text{ AgCl}}{1\text{ AgNO}_3}\text{ = 0.314 mol AgCl}[/tex]

For NaCl

[tex]14.19g\text{ NaCl x }\frac{1\text{ mol NaCl}}{58,44g}\text{ x }\frac{1\text{ AgCl}}{1\text{ NaCl}}\text{ = 0.243 mol AgCl}[/tex]

Therefore the limiting reagent is NaCl

Step 2: Calculate the mass of AgCl

[tex]0.243\text{ mol AgCl x }\frac{143,32g\text{ AgCl}}{1mol\text{ AgCl}}\text{ = 34.8 g AgCl}[/tex]

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