ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: eeps on February 06, 2011, 08:41:28 pm
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I don't know what's the go with having one single Question Thread, though I need a few questions answered! I'll use this thread throughout the year, if I have any questions.
1. In an analysis of a gas, the mass of 5.00 L at STP was found to be 14.3 g. Which one of the following is the gas most likely to be?
A) Cl2 B) H2 C) CO2 D) SO2
2. Consider the following equation; 2Al(s) + Cr2O3(s) -> Al2O3(s) + 2Cr(s). What is the theoretical yield of chromium that can be produced by the reaction of 40.0 g of Cr2O3 with 8.00 g of aluminium?
A) 7.70 g B) 15.4 g C) 27.3 g D) 30.8 g
Also, I suck at using LaTeX.
Thanks to anyone who can help! =]
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1. pv=nrt...
find n ..., n=0.223
n=mass/molar mass, rearrange to find molar mass...
molar mass = 64.05..
hence D is the answer SO2
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Thanks! =]
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2. Only hints: Try finding which is in excess (M is known and m is known, find n), then work from there
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for Question 2, I got B..
40/(51.996x2+16x3) [mol of Cr2O3]is greater than (8/26.982)/2 (mol of Al) so Al is the limiting reagent, so the number of mole of Cr produced shall be the same as 8/26.982 so its B....
not very confident here ^^ but i assume you have the answer with you?
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for Question 2, I got B..
40/(51.996x2+16x3) [mol of Cr2O3]is greater than (8/26.982)/2 (mol of Al) so Al is the limiting reagent, so the number of mole of Cr produced shall be the same as 8/26.982 so its B....
not very confident here ^^ but i assume you have the answer with you?
haha ..that's correct~~
confirmed
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for Question 2, I got B..
40/(51.996x2+16x3) [mol of Cr2O3]is greater than 8/26.982 (mol of Al) so Al is the limiting reagent, so the number of mole of Cr produced shall be the same as 8/26.982 so its B....
not very confident here ^^ but i assume you have the answer with you?
Nope, I don't have the answers. Though I got B as well, so I'll assume I'm right. Thanks.