Final answer:
The reaction between barium chloride and sulfuric acid produces barium sulfate and hydrochloric acid. By calculating the theoretical yield of barium sulfate and comparing it to the experimental yield, you can determine the relative error of the reaction. In this case, the relative error is 9.59%.
Step-by-step explanation:
The reaction between barium chloride (BaCl2) and sulfuric acid (H2SO4) produces barium sulfate (BaSO4) and hydrochloric acid (HCl). The balanced equation for the reaction is:
BaCl2 + H2SO4 → BaSO4 + 2HCl
To calculate the relative error of the reaction, we need to find the theoretical yield of barium sulfate. The molar mass of BaCl2 is 208.23 g/mol and the molar mass of BaSO4 is 233.39 g/mol. Using the given mass of sulfuric acid solution (9.8 g) and the molar mass of H2SO4 (98.09 g/mol), we can calculate the number of moles of H2SO4 used:
moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4 = 9.8 g / 98.09 g/mol = 0.100 moles
From the balanced equation, we can see that 1 mole of H2SO4 reacts with 1 mole of BaCl2 to produce 1 mole of BaSO4. Therefore, the theoretical yield of BaSO4 is also 0.100 moles. To convert this to grams:
mass of BaSO4 = moles of BaSO4 * molar mass of BaSO4 = 0.100 moles * 233.39 g/mol = 23.339 grams
The relative error of the reaction can be calculated using the formula:
relative error = (|experimental yield - theoretical yield| / theoretical yield) * 100%
Given the mass of the white precipitate formed (21.1 grams), the experimental yield of BaSO4 can be calculated:
experimental yield = 21.1 grams
Now we can calculate the relative error:
relative error = (|21.1 g - 23.339 g| / 23.339 g) * 100% = 9.59%