Barium carbonate and nitric acid:
BaCO3 (s) + 2HNO3 (aq.) --> Ba2+ (aq.) + 2(NO3-) (aq.) + CO2 (g) + H2O (l)
If 25.0mLs of 0.150M nitric acid was used, how many grams of barium carbonate would be required for complete reaction?
jlasn
To determine the grams of barium carbonate required for complete reaction, we can use the balanced equation and stoichiometry.
First, let's calculate the amount of HNO3 used in moles:
Moles of HNO3 = volume (L) × concentration (M)
Moles of HNO3 = 0.025 L × 0.150 mol/L
Moles of HNO3 = 0.00375 mol
According to the balanced equation, the stoichiometric ratio between HNO3 and BaCO3 is 2:1. This means that 2 moles of HNO3 react with 1 mole of BaCO3.
Using this ratio, we can determine the moles of BaCO3 needed:
Moles of BaCO3 = 0.00375 mol NO3- × (1 mol BaCO3 / 2 mol NO3-)
Moles of BaCO3 = 0.001875 mol
Finally, we can convert moles of BaCO3 to grams using its molar mass:
Grams of BaCO3 = moles of BaCO3 × molar mass of BaCO3
Grams of BaCO3 = 0.001875 mol × (137.33 g/mol + 12.01 g/mol + 3(16.00 g/mol))
Grams of BaCO3 = 0.001875 mol × 197.34 g/mol
Grams of BaCO3 ≈ 0.3697 g
Therefore, approximately 0.3697 grams of barium carbonate would be required for the complete reaction.
To find the grams of barium carbonate required for complete reaction, we need to use the balanced equation and molarity of the nitric acid.
1. Write down the balanced equation:
BaCO3 (s) + 2HNO3 (aq.) --> Ba2+ (aq.) + 2(NO3-) (aq.) + CO2 (g) + H2O (l)
2. Determine the mole-to-mole ratio between nitric acid (HNO3) and barium carbonate (BaCO3) from the balanced equation. In this case, the ratio is 1 mol HNO3 to 1 mol BaCO3.
3. Calculate the moles of nitric acid used:
Molarity (M) = moles/volume (L)
0.150 M = moles/0.0250 L
moles = 0.150 M * 0.0250 L = 0.00375 mol
4. Since the mole-to-mole ratio is 1:1, the moles of barium carbonate required is also 0.00375 mol.
5. Find the molar mass of barium carbonate (BaCO3):
Ba: 1 * 137.33 g/mol = 137.33 g/mol
C: 1 * 12.01 g/mol = 12.01 g/mol
O: 3 * 16.00 g/mol = 48.00 g/mol
Total molar mass = 137.33 + 12.01 + 48.00 = 197.34 g/mol
6. Convert moles of barium carbonate to grams:
Grams = moles * molar mass
Grams = 0.00375 mol * 197.34 g/mol = 0.738 g
Therefore, 0.738 grams of barium carbonate would be required for complete reaction with 25.0 mL of 0.150 M nitric acid.