Chemistry Question...?
Answer:
You have 0.8 L x 0.04 M = 0.032 mol Ba(OH)2
You have 0.4 L x 0.2 M = 0.08 mol HCl.
Ba(OH)2 is limiting. (Plenty of HCl to convert all to BaCl2 and some extra. The previous answer assumed that there was an exact amount of HCl and based the calculations on that, which is exactly the kind of trap that your professor hopes you will fall into when you take the exam.)
Your total volume is 1.2 L not counting the water liberated in the reaction.
Water liberated in the reaction is 2 x 0.032 mol x 18 g/mol = 1.15 g, which is not significant here because you only have one significant figure on your volumes. For extremely precise work this could matter.
0.032 mol BaCl2 / 1.2 L = 0.027 M
Hello
your answer is 0. 033 M BaCl2
every 2 mole of HCl neutralize with 1 mole of Ba(OH)2
then the total volume of the neutralization is 1.2 liter
then 0.08 mole HCl (0.4 lite*0.2 M) divided to 2 to get the mole consumption of Ba(OH)2 .it becomes 0.04 mole then u should divide the mole of BaCl2 to the total volume is (400 ml+800ml=1200ml=1.2 Liter) it becomes 0.03333333
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