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12345 [234]
2 years ago
7

How much aluminum nitrate can be found when 21g of aluminum react with lead (II) nitrate? Answer in units of mol.

Chemistry
1 answer:
Bond [772]2 years ago
7 0

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.

Learn more moles at: brainly.com/question/13314627

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...<em>To make exactly 100.0mL of solution...</em>

Molar concentration is defined as the amount of moles of a solute (In this case, nitrate ion, NO₃⁻) in 1 L of solution.

To solve this question we need to convert the mass of Fe(NO₃)₃ to moles. As 1 mole of Fe(NO₃)₃ contains 3 moles of nitrate ion we can find moles of nitrate ion in 100.0mL of solution, and we can solve the amount of moles per liter:

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