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HCl + AgNO3----------- AgCl + HNO3

To get our limiting reagent that will be used for the reaction:

Using n=CV for Hcl

C=1mol/L V=50/1000= 0.05L

n=1×0.05=0.05moles

Therefore using a mole-mass relationship

If 1mole of HCl react with 169.9g of AgNO3

Then the the mass of AgNO3 that will react with 0.05moles of HCl= 0.05×169.9/1 =8.495g

Therefore AgNO3 is in excess since we have 53g of AgNO3

Then HCl is the limiting reagent

Molar mass of Agcl is 143.3

So if 1 mole of HCl react to give 143.3g of Agcl

Then amount of AgCl that will be produced if 0.05moles of HCl react is =0.05×143.3/1=7.165g

The answer is 7.165g

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