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If 14.5 grams of silver nitrate (molar mass =169.88g/mol) reacts with aluminum chloride. How many grams of aluminum nitrate (molar mass = 213.01g/mol) are formed

If 14.5 grams of silver nitrate (molar mass =169.88g/mol) reacts with aluminum chloride. How many grams of aluminum nitrate (molar mass = 213.01g/mol) are formed
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If 14.5 grams of silver nitrate (molar mass =169.88g/mol) reacts with aluminum chloride. How many grams of aluminum nitrate (molar mass = 213.01g/mol) are formed

User Shlomit
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1 Answer

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3AgNO3 + AlCl3 ----> Al(NO3)3 + 3AgCl


14.5 * ( (1mol AgNO3)/(169.88g AgNO3) ) * ( (1mol AlNO3)/(3mol AgNO3)) * ( (213.01g AlNO3)/(1mol AlNO3) ) = 6.06gAl(NO3)3

The answer is 6.06g of Al(NO3)3 is formed.
User Alejo Bernardin
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