急! F5 chem 兩條難題

1.
All of A, B, C and D are NOT the correct answer.

I think there is a mistake in setting the question:
1.5 M HNO3­ is a moderately strong oxidizing agent. Therefore, the reaction between Mg and HNO3 gives nitrogen monoxide gas, but not H2.

3Mg + 8HNO3 → 3Mg(NO3)2­ + 4H2O + 2NO
No. of moles of Mg added = 3/24.3 = 0.123 mol
No. of moles of HNO3 added = 1.5 x (50/1000) = 0.075 mol
When all Mg reacts, HNO3 needed = 0.123 x (8/3) = 0.328 mol
Hence, Mg is in excess, and HNO3 is the limiting reactant.
No. of moles of Mg(NO3)2 formed = 0.075 x (3/8) = 0.0281 mol

Original no. of moles of Mg(NO3)2 = 2 x (100/1000) = 0.2 mol
Total no. of moles of Mg(NO3)2 = 0.2 + 0.0281 = 0.2281 mol
Volume of the final solution = 100 + 50 = 150 cm3 = 0.15 dm3
Molarity of Mg(NO3)2 = 0.2281/0.15 = 0.152 M

The answer is close to C, but answer C is calculated based on the incorrect reaction in which hydrogen is formed.


2.
The answer is C.

Let there are h mol of MCO3 and k mol of NHCO3.

MCO3 + H2A → MA + H2O + CO2 ...... (*)
2NHCO3 + H2A → N­2A + 2H2O + 2CO2 ...... (#)

Total mol. of H2A used = Mol. of H2A in (*) + Mol. of H2A in (#)
0.05 x (110/1000) = h + k(1/2)
2h + k = 0.011 ...... (1)

Total mol. of CO2 formed = Mol. of CO2 in (*) + Mol. of CO2 in (#)
168/24000 = h + k
h + k = 0.007 ...... (2)

(1) - (2): h = 0.004
2(2) - (1): k = 0.003

Mole ratio MCO3 : NHCO3 = h : k = 0.004 : 0.003 = 4 : 3
=

i just can calculate no.1 out...hope u do not mind that..

FULL ANSWER::

no of mole of Mg(NO3)2 :: 2 x (100/1000) = 0.2mol
no of mole of HNO3:: 1.5x(50/1000)= 0.075mol

By this chemical equation,
(when HNO3 is added to the solution)

Mg(s)+2HNO3(aq) -------> Mg(NO3)2 (aq)+H2

the no. of mol of Mg(NO3)2 : 0.075/2 = 0.0375 mol

the total no of mol of Mg(NO3)2 :: 0.0375+0.2 = 0.2375mol

the total volume in dm3 = (100+50)/1000 = 0.15dm3

the molarity of Mg(NO3)2 : 0.2375/0.15 = 1.58 M (cor. to 3 sig. fig.)



ANS:: C




LET'S FIGHT FOR CE!! + OIL =]