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Neporo4naja [7]
4 years ago
14

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

ar mass = 213.01g/mol) are formed
Chemistry
1 answer:
PtichkaEL [24]4 years ago
7 0
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.
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From the given information, m = 0.905 g, initial temperature (room temperature) = 21°C , final temperature = 175°C, Q = ?

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