Question

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

Answer 1

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

$14.5 * ( \frac{1mol AgNO3}{169.88g AgNO3} ) * ( \frac{1mol AlNO3}{3mol AgNO3}) * ( \frac{213.01g AlNO3}{1mol AlNO3} ) = 6.06gAl(NO3)3$

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