Given:
<span>BaO2(s) + 2 HCl(aq) => H2O2(aq) + BaCl2(aq)
</span>
Mass of barium peroxide = 1.49 g Volume of HCl solution = 25.1 mLConcentration of HCl solution = 0.0279 g HCl per mL
Solution: To determine the amount of hydrogen peroxide that would be produced, we use need to determine which is the limiting reactant and use the initial amount of this reactant for the calculations.
0.0279 g HCl / mL ( 25.1 mL ) ( 1 mol HCl / 36.46 g HCl) ( 1 mol BaO2 / 2 mol HCl ) = 0.0096 mol BaO2 needed
1.49 g BaO2 ( 1 mol / 169.3 g BaO2 ) ( 2 mol HCl / 1 mol BaO2 ) = 0.0176 mol HCl needed
Therefore, the limiting reactant would be barium peroxide since it is consumed completely first in the reaction.
1.49 g BaO2 ( 1 mol BaO2 / 169.3 g BaO2 ) ( 1 mol H2O2 / 1 mol BaO2 ) ( 34.02 g H2O2 / 1 mol H2O2 ) = 0.299 g H2O2
Iron fillings can be separated as they are magnetic. Use a magnet to separate the iron fillings away.
Salt can be separated away as they are soluble in water.
Answer:
The volume of a sample of gas (2.49 g) was 752 mL at 1.98 atm and 62∘C 62 ∘ C .
Answer: barium oxide
Explanation: The barium will give some of it's electrons up to the oxygen, and then they will both reach the stability of a noble gas. Then, they will both combinate to barium oxide, so the answer is barium oxide.
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