If 25.0 mLs of 0.150M nitric acid was used, how many grams of barium carbonate would be required for complete reaction?
This is the same type of problem but it isn't a limiting reagent problem.
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To find the number of grams of barium carbonate required for complete reaction, you need to use the balanced equation for the reaction between nitric acid (HNO3) and barium carbonate (BaCO3).
The balanced equation for the reaction is:
2 HNO3 + BaCO3 → Ba(NO3)2 + CO2 + H2O
From the balanced equation, you can see that 2 moles of nitric acid react with 1 mole of barium carbonate.
To find the number of moles of nitric acid used, you can use the formula:
moles = concentration × volume (in liters)
In this case, the concentration of nitric acid is 0.150 M (moles per liter) and the volume is 25.0 mL (0.0250 L). Therefore, the number of moles of nitric acid used is:
moles of HNO3 = 0.150 M × 0.0250 L = 0.00375 moles
Since the mole ratio between nitric acid (HNO3) and barium carbonate (BaCO3) is 2:1, you know that:
moles of HNO3 : moles of BaCO3 = 2 : 1
Therefore, the number of moles of barium carbonate required for complete reaction is:
moles of BaCO3 = 0.00375 moles ÷ 2 = 0.001875 moles
Finally, to find the mass of barium carbonate required, you need to use the molar mass of BaCO3. The molar mass of BaCO3 is:
BaCO3: 1 × Ba = 137.33 g/mol
1 × C = 12.01 g/mol
3 × O = 3 × 16.00 g/mol = 48.00 g/mol
Therefore, the molar mass of BaCO3 is:
Molar mass of BaCO3 = 137.33 g/mol + 12.01 g/mol + 48.00 g/mol = 197.34 g/mol
Using this molar mass, the mass of barium carbonate required for complete reaction is:
mass of BaCO3 = moles of BaCO3 × molar mass of BaCO3
= 0.001875 moles × 197.34 g/mol
So, the amount of barium carbonate required for complete reaction is:
mass of BaCO3 = 0.370 g (rounded to three decimal places)