Question

For the following reaction, 23.8 grams of hydrochloric acid are allowed to react with 60.2 grams of barium hydroxide. hydrochloric acid(aq)+ barium hydroxide(aq) - barium chloride(aq) + water(l) What is the maximum amount of barium chloride that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer 6 more group attempts remaining Retry Entire Group

Answer 1

HCl + Ba(OH) Back + H2O After balancing 2HCl + Ba(OH) Bach, + 2H₂O According a balanced stoichiometric equation, a moles of Hel reacts with Imole of BacoHz to give Imole of Bach. so Hel G BaCOH) are should be seact in 2:1 ratio. Let us calculate the no. of nolex up нс! Во (он), Number of noks of Hel: Molecules weight of Hol- 36.460 14 golmal weight of Hd = 23.89

8 of molex of Ha- Weight of Hol 2 Molecules weight of HCL 2.3.89 36.460948 Imoll = 0.65275333 mol Number of roles of BacoH₂: Molecules weight of Balot2 = 171.34 g/mol weight of BacoH = 60.2g. Number of moles of BacoH), Weight of Barot), op Bacot), Moleculeg weight 60. 2g 171.34 g/mol 0. 35/3481966 mol

The ratio of Hel G BalOH)₂ 0.652 75333 mol : 0.353481966 mol Devide with least number 0.652.75333 me 0.3513481966 mol comel . 0.3513481966 mol 0.3513481966 mol 1.8578530823 : 1 it But the ratio op HCIE BacoH), 1.857 8530823 : 1 to But required satio equation is should be according 2:1 less amant the present to deact in with whole of so Hol is that required the Bacott2. so Hcl it Bacoth, reagent limiting the ix penent in excess.

Now let us answer the quextion (4 Maximum amount of Bach, that can be formed: since limiting reagent is Her so the maximum amount of Bach that can be formed is equal to depende on the amount of Hd. According to equation 2 nolex of Hcl to can form Imole of - Bach. then 0.65275333moles of HCl cantom ? 2moler optic - Imole of Bach 0.652 753 33 mole, Phd x 9 x 2molex of Hel=Imole of Bach X 0.65 2.75333 molesale HCO

Imolep Bacle X 0.65275333 mole-pHCH 2 notes of Hel 1 x 0.65275333 male or Bade 9 = 0.326 37 6665 males op Bacle we know that w no. of moles Weight moleruby weight leight = norep moles x molecular weight Molecules weight of Bach = 208. 23g/mol noop molex - Radly = 0.32.6376665 molas & Weight up Bacho = 0.32 637 6665 moles x 208.23 g/mol - 67.9614129539 . Maximum amount of Back that can be formed = 67.96141295398 - 67.969

bt? The formula of limiting reagent : formula As we discussed earlier the of limiting reagent is Hol! Hydrochloric acid. Amount op excess reagent remains apter the reaction is complete we know that excess reagent in BacOH)2 According 2moles op to equation Hal scocty with Imole op BacoH than 0.652_75333 mol op. Hal Beact with amoles of Hol a Imole of Bacon) 0.65 275 333 mol op HCl -> 2 => ? mole op Hc) = mole of Baco), X 0.65 2.75333 maler op HCl

7 - Imole op. Ba (CH), X 0.65 275333 moles op HCL 2 mole, of Hol 1X0.65275333 molex op Balott), 0.32637 6665 moles op Baco), so to react op HCl with 0.65275333 nolex 0.32637 6665 moles of Ba(OH2 is enough. But we have 0.35/34 81966 molex op BaCOH)2 so the number of molex remaining - Total number of moles - No. of molex reacted 0.35/3481966.cole - 0.326376665 moler = 0.249715 316 moles

We know that (8 molecular weight weight - No. of molex x No. of molex of Ba(OH)2 remaining, -0.0249715316 molen Moleculen weight op Balo), = 11.34 g/mol 2. Amount op BacoH), semaining = 0.024971531 G molex x 171.349/mol - 4,27 862 222 43 g - 4.28g Amount & BalOH semaning after reaction is complete is 4.289