Considering the neutralization reaction when nitric acid reacts with solid iron(II)carbonate. What is the minimum volume of 2.00 M HNO3 necessary to react completely with 1.06 kg of solid iron(II) carbonate, FeCO3 ?
FeCO3 + 2HNO3 ==> Fe(NO3)2 + H2O + CO2
1.06 kg Fe = 1060 grams.
mols FeCO3 = grams/molar mass.
Using the coefficients in the balanced equation, convert mols FeCO3 to mols HNO3.
Then mols HNO3 = M x L. You know mols and M, solve for L and convert to whatever unit (mL for example) if needed.
To find the minimum volume of 2.00 M HNO3 necessary to react completely with 1.06 kg of solid FeCO3, we need to determine the number of moles of FeCO3, then use stoichiometry to find the number of moles of HNO3 required, and finally convert that to volume.
Step 1: Find the number of moles of FeCO3
The molar mass of FeCO3 is:
Fe = 55.845 g/mol
C = 12.011 g/mol
O = 15.999 g/mol (three oxygen atoms)
So, the molar mass of FeCO3 is: 55.845 + 12.011 + (15.999 × 3) = 115.853 g/mol
To find the number of moles, we divide the mass of FeCO3 by its molar mass:
moles of FeCO3 = mass of FeCO3 / molar mass of FeCO3
moles of FeCO3 = 1.06 kg × (1000 g / 1 kg) / 115.853 g/mol
Step 2: Determine the stoichiometry between FeCO3 and HNO3
From the balanced chemical equation, we know that the molar ratio between FeCO3 and HNO3 is 1:2 (1 mole of FeCO3 reacts with 2 moles of HNO3).
Step 3: Calculate the moles of HNO3 required
moles of HNO3 = moles of FeCO3 × 2
Step 4: Convert moles of HNO3 to volume
To calculate the volume, we need to use the molarity of HNO3. The equation for molarity is:
Molarity = moles / volume (in liters)
Rearranging the equation, we get:
Volume (in liters) = moles / molarity
Let's calculate the moles of FeCO3 and then find the volume of HNO3 required.
moles of FeCO3 = 1.06 kg × (1000 g / 1 kg) / 115.853 g/mol
moles of FeCO3 = 9.152 mol
moles of HNO3 = 9.152 mol × 2
moles of HNO3 = 18.304 mol
Volume of HNO3 = moles of HNO3 / molarity
Volume of HNO3 = 18.304 mol / 2.00 mol/L
Volume of HNO3 = 9.152 L
Therefore, the minimum volume of 2.00 M HNO3 necessary to react completely with 1.06 kg of solid FeCO3 is 9.152 liters.
To determine the minimum volume of 2.00 M HNO3 required to react completely with 1.06 kg of solid iron(II) carbonate, FeCO3, we need to follow these steps:
Step 1: Write the balanced chemical equation for the neutralization reaction.
The balanced chemical equation between nitric acid (HNO3) and iron(II) carbonate (FeCO3) is:
2 HNO3 + FeCO3 -> Fe(NO3)2 + CO2 + H2O
Step 2: Calculate the molar mass of FeCO3.
Fe = 55.85 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
Molar Mass of FeCO3 = (55.85 g/mol) + (12.01 g/mol) + 3 * (16.00 g/mol)
Molar Mass of FeCO3 = 115.87 g/mol
Step 3: Convert the mass of FeCO3 to moles.
moles of FeCO3 = mass of FeCO3 / molar mass of FeCO3
moles of FeCO3 = 1.06 kg / (115.87 g/mol)
moles of FeCO3 = 9.15 mol
Step 4: Use the balanced equation to determine the mole ratio between HNO3 and FeCO3.
From the balanced equation, we see that it takes 2 moles of HNO3 to react with 1 mole of FeCO3.
Step 5: Calculate the number of moles of HNO3 required.
moles of HNO3 = 2 * moles of FeCO3
moles of HNO3 = 2 * 9.15 mol
moles of HNO3 = 18.3 mol
Step 6: Calculate the volume of 2.00 M HNO3 required.
volume (in liters) of HNO3 = moles of HNO3 / Molarity
volume (in liters) of HNO3 = 18.3 mol / 2.00 M
volume (in liters) of HNO3 = 9.15 L
Therefore, the minimum volume of 2.00 M HNO3 necessary to react completely with 1.06 kg of solid iron(II) carbonate, FeCO3, is 9.15 liters.