0.77 moles of aluminum nitrate can be found when 21g of aluminum react with lead (II) nitrate.
<h3>Equation of reaction</h3>
- Al + Pb(NO₃)₂ ---> Al(NO₃)₃ + Pb (s)
From the equation of reaction:
1 mole of aluminum reacts with 1 mole of lead nitrate to produce 1 mole of aluminum nitrate
1 mole aluminum has a mass of 27 g
Number of moles of aluminum in 21 g = 21/27 =0.77 moles
0.77 moles of Aluminum produces 0.77 moles of aluminum nitrate
Therefore, 0.77 moles of aluminum nitrate can be found when 21g of aluminum react with lead (II) nitrate.
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