1. eli123

2. eli123

@phi

3. eli123

@robtobey

4. NoelGreco

For someone to comment, he's have to know more about the lab.

5. aaronq

the moles you can simply find by dividing the masses by their molar mass... for the volume of H2 formed, you need to write the equation of the reaction in your exp and work with that.

6. eli123

@aaronq so for magnesium it would be 24.31/0.035?

7. aaronq

the other way around

8. eli123

oh ok and for h2? @aaronq

9. aaronq

you have to find the volume first

10. eli123

is the volume of hydrogen gas produced the same as teh volume of liquid displaced? @aaronq

11. aaronq

12. aaronq

yes

13. eli123

no i didnt

14. eli123

so the volume divided by the mass of h2 will give me the moles of h2?

15. aaronq

well for ideal gases you can just use the fact that 1 mole = 22.414 L at STP

16. aaronq

otherwise you have to know the density of the gas

17. eli123

therefore, how would the equation look?

18. aaronq

so for ideal gases, moles = volume of gas/(22.414 L/mol)

19. aaronq

make sure you have the same units

20. eli123

oooh

21. eli123

@aaronq for the moles of magnesium igot 0.001 for all the trials, but my friend wrote it in dimensional analysis . whats betteR?

22. aaronq

it's the same thing, since it's a gas, the volume is proportional to the amount of moles

23. eli123

ok and how would i find the ratio of volume displaced in mL to moles h2 gas produced? @aaronq

24. aaronq

a ratio is just a division volume of gas/moles of gas

25. eli123

so volume of h2/moles of h2?

26. eli123

@aaronq

27. aaronq

yeah

28. eli123

@aaronq i get a huge number

29. eli123

37.00ml/0.002

30. aaronq

yep, thats what the want

31. eli123

18500?

32. aaronq

they actually want the volume displaced/molesof H2 which is the same thing

33. aaronq

mL/mol

34. eli123

so its right?

35. eli123

@aaronq

36. aaronq

to what they're asking you, yes

37. eli123

but its such a big number? @aaronq

38. aaronq

yep thats not that big of a number, remember it's in mL ..

39. eli123

18.5 L DAMM

40. aaronq

lol that's not a lot, it really should be around 22.414 L

41. eli123

ARE YOU POSITIVE?

42. aaronq

well unless they want you to calculate using Mg (s) + HCl -> H2 + MgCl (aq)

43. eli123

how would that be?

44. aaronq

this would be the theoretical way you balance the equation and use diminutional analysis, but really your answer is gonna be like off by very little, also, because of rounding

45. aaronq

did they give you the concentration of the acid?

46. eli123

no

47. eli123

all we had to do was place a piece of magnesium inside a buret that contained hcl and this was then converted and all the hcl migrated down to the bottom of the buret

48. eli123

basically the magnesium was reacting with the hcl

49. eli123

and after that we had to find the values in that table

50. aaronq

hm well they really wouldn't need to tell you, do you know what a limiting reagent is?

51. eli123

limiting reactant?

52. aaronq

yep, it's the same thing. you could use the number of moles of that and translate it to moles of H2 produced 2Mg (s) + 2HCl -> H2 + 2MgCl (aq) 0.001439 moles of Mg/2 = x moles of H2/1 moles of H2 = 0.0028794734677088 thats the theoretical yield

53. eli123

well that makes more sense because thats what we are currently learning in class haha

54. aaronq

oh wait sorry i multiplied, it should be moles of H2 = 0.0007198683669272

55. aaronq

haha oh true, well maybe let me know next time? haha

56. eli123

sorry!!!

57. eli123

so in other words, to find lets say trial #1

58. eli123

where the volume of hydrgen gas produced was 37.00mL and the moles of h2 gas produced were 0.002 mol

59. aaronq

you would convert the mass of Mg to moles then using the balanced equation, build a relationship between H2 and Mg using their respective coefficients.. this just means: moles of Mg/coefficient=moles of H2/coefficient

60. eli123

oh my god

61. eli123

moles of mg that reacted?

62. eli123

coefficient is 2?

63. aaronq

yeah because thats your limiting reactant..

64. aaronq

for Mg is 2 for H2 is 1

65. aaronq

wait no i balanced it wrong because it's Mg + 2HCl -> MgCl2 + H2

66. aaronq

so it's coefficient is 1 for both

67. aaronq

sorry, i'm like lost today idk why

68. eli123

i dont get it

69. aaronq

Mg (s) + 2HCl -> H2 + MgCl2 0.001439 moles of Mg/1 = x moles of H2/1 moles of H2 = 0.001439 thats the theoretical yield which is close to what you would get if you just divided the volume of gas produced by 22.414 L

70. eli123

thats the ratio?

71. aaronq

yep the ratio is 1:1

72. eli123

but what value would i put?

73. aaronq

i would go with the theoretical since thats what you're doing in class

74. eli123

0.001439?

75. aaronq

yeah, even round up to 0.0015

76. aaronq

or down to 0.0014

77. eli123

ok wait where did the 0.001439 come from>?

78. aaronq

its the number of moles of magnesium

79. eli123

ooooh yes thats my 0.001

80. aaronq

haha yep, i mean round to whatever you think it's best, or whatever they're teaching you at school

81. eli123

but for the ratio isnt moles of h2 divided by moles of magnesium? so 0.002h2/0.001mg

82. aaronq

which ratio?

83. eli123

ratio of volume displaced to moles h2 gas prodced

84. aaronq

well the amount of moles of H2 produced is equal to the moles of Mg that reacted according to the stoichiometric ratios of the equation... so to answer your question it would be 37 mL/0.001 moles.. which is what you had before, if i remember correctly

85. eli123

37 being the volume of hydrogen gas produced and 0.001 moles of magnesium reacted?

86. aaronq

well, technically yes, but the question is asking for the volume displaced divided by the moles of H2 produced..which are the same values

87. eli123

oh so volume of liquid displaced divided by the moles of h2

88. eli123

and thats the ratio?

89. eli123

wow i feel stupid

90. aaronq

that they're asking you for in the question yes. They're basically seeing how close that value is to the ideal gas constant volume of 1 mole =22.414 L, since 37mL/0.0014 mol = 26428.5714285714285714 mL/mol 26.428 L/mol so it's close if you use more than 0.001, which i think you should do lol

91. eli123

oh my i get the big number again hahaha

92. aaronq

haha don't feel stupid

93. aaronq

94. eli123

0.0014 is in the correct significant figures? i keep using 0.001

95. aaronq

hm idk though if you don't use that your error is pretty large

96. eli123

exactly

97. eli123

but look when i divide 0.035/ 24.31

98. aaronq

0.0014397367338544

99. eli123

and the right sig figures are four so you round it to 0.001?

100. aaronq

zeros dont count though

101. aaronq

well zeros on the left side

102. eli123

oh..

103. eli123

still..

104. eli123

so now its 3 sig figures?

105. aaronq

i would use 2 sig digits b/c 0.035 .. and have 0.0014

106. eli123

so only 14 are significant

107. aaronq

yes

108. eli123