How many grams of silver chloride can be produced if you start with 4.62 grams of barium chloride ? 2agno3 + bacl2 —> 2agcl + ba(no3) show work.

First, we need to get the moles of  BaCl2:

When we have the mass of barium chloride = 4.62 g  

and the molar mass of BaCl2 = 208.23 g /mol

so by using this values, we can get the moles of BaCl2:

moles of BaCl2 = mass / molar mass
       
                          = 4.62 g / 208.23 g/mol
                          = 0.022 mol

according to the balanced equation of the reaction:

2AgNO3 + BaCl2 → 2 AgCl + Ba(NO3)2

so you can see from this equation that the molar ratio between

BaCl2 & AgCl = 1 : 2 

∴ moles AgCl = moles BaCl2 * 2
 
                       = 0.022 * 2 = 0.044 moles 

now, we have the no.of moles of AgCl so by using the molar mass of AgCl we can get the mass of AgCl:

Mass AgCl = moles AgCl * molar mass
    
                   = 0.044g * 143.3cg/mol 

                   = 6.3 g 

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