Calculate the volume (in mL) of 0.627 M HNO3 needed to react completely with 8.12 g of BaCO3 in a gas-forming reaction
To calculate the volume of 0.627 M HNO3 needed to react completely with 8.12 g of BaCO3, we need to follow these steps:
1. Determine the balanced equation for the reaction between HNO3 and BaCO3.
The balanced equation is:
2 HNO3 + BaCO3 → Ba(NO3)2 + H2O + CO2
2. Find the molar mass of BaCO3.
The molar mass of BaCO3 is:
BaCO3 = (1 × Ba) + (1 × C) + (3 × O) = 137.33 g/mol
3. Convert the given mass of BaCO3 to moles.
Using the formula:
moles = mass / molar mass
moles of BaCO3 = 8.12 g / 137.33 g/mol = 0.0592 mol
4. Determine the stoichiometry between HNO3 and BaCO3.
From the balanced equation:
2 moles of HNO3 react with 1 mole of BaCO3
Therefore, the moles of HNO3 needed = 2 × moles of BaCO3 = 2 × 0.0592 mol = 0.1184 mol
5. Calculate the volume of 0.627 M HNO3 needed.
Using the formula:
moles = concentration × volume
volume = moles / concentration
volume of 0.627 M HNO3 = 0.1184 mol / 0.627 mol/L = 0.189 L = 189 mL
Therefore, the volume of 0.627 M HNO3 needed to react completely with 8.12 g of BaCO3 is 189 mL.