How much aluminum nitrate can be found
when 12 g of aluminum react with lead(II)
nitrate?
Answer in units of mol
2Al + 3Pb(NO3)2 ==> 2Al(NO3)3 + 3Pb
1. Convert 12 g Al to moles. moles = g/molar mass
2. Using the coefficients in the balanced equation, convert mol Al to moles Al(NO3)3
3. Convert moles Al(NO3)3 to g. g = moles x molar mass.
To determine the amount of aluminum nitrate formed when 12 g of aluminum reacts with lead(II) nitrate, we need to use the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between aluminum and lead(II) nitrate is:
2 Al + 3 Pb(NO3)2 -> 2 Al(NO3)3 + 3 Pb
From the equation, we can see that the molar ratio between aluminum and aluminum nitrate is 2:2. This means that for every 2 moles of aluminum reacting, we will get 2 moles of aluminum nitrate produced.
To find the number of moles of aluminum nitrate, we first need to convert the mass of aluminum into moles. The molar mass of aluminum (Al) is approximately 26.98 g/mol.
Number of moles of aluminum = mass / molar mass
Number of moles of aluminum = 12 g / 26.98 g/mol
Number of moles of aluminum = 0.445 mol (rounded to three decimal places)
Since the molar ratio between aluminum and aluminum nitrate is 2:2, we can conclude that the number of moles of aluminum nitrate formed is also 0.445 mol.
Therefore, when 12 g of aluminum reacts with lead(II) nitrate, the amount of aluminum nitrate that can be formed is 0.445 mol.
To find the amount of aluminum nitrate produced when 12 g of aluminum reacts with lead(II) nitrate, you need to follow these steps:
Step 1: Write the balanced chemical equation.
2Al + 3Pb(NO3)2 → 2Al(NO3)3 + 3Pb
Step 2: Calculate the molar mass of aluminum (Al).
The molar mass of aluminum (Al) is 26.98 g/mol.
Step 3: Convert the given mass of aluminum (12 g) to moles.
Moles of aluminum (Al) = Mass of aluminum (Al) / Molar mass of aluminum (Al)
Moles of aluminum (Al) = 12 g / 26.98 g/mol
Moles of aluminum (Al) ≈ 0.445 mol (rounded to three decimal places)
Step 4: Use the mole ratio from the balanced chemical equation to find the moles of aluminum nitrate (Al(NO3)3).
From the balanced chemical equation, the mole ratio between aluminum (Al) and aluminum nitrate (Al(NO3)3) is 2:2.
Moles of aluminum nitrate (Al(NO3)3) = Moles of aluminum (Al)
Moles of aluminum nitrate (Al(NO3)3) = 0.445 mol
Step 5: Round the moles of aluminum nitrate (Al(NO3)3) to the appropriate number of decimal places according to the significant figures in the question.
Since 12 g of aluminum is given, the answer should also have three decimal places.
Therefore, the amount of aluminum nitrate (Al(NO3)3) produced when 12 g of aluminum reacts with lead(II) nitrate is approximately 0.445 mol.