please help me with these practice test problems it will help me study for the u
ID: 1006212 • Letter: P
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please help me with these practice test problems it will help me study for the upcoming test
thank you and much appreciated
d. How much of the ER remains when the reaction goes to completion? g. What is the percent composition of a. magnesium nitrate? b. aluminum sulfide? 10. A chemist wants to prepare a 0.75 M HCI solution using a 15.8 M stock solution. How many mL of the stock solution are needed to make 3.5 L of 0.75 M HCI? 11. How many mL of 0.250 M KMno4 are needed to react with 3.36 g of Feso4? 10 Feso4 2 KMno4 8 H2SO 5 Fe2(SO4)3 2 MnSO4 K2SO4 8 H20 4 12. What is the concentration of all species in: a. 0.25 M barium nitrate? b. 0.50 M cesium chloride? c. 0.15 M Al(OH)3? 13. You determine using stoichiometry that a reaction produces 5.9 g of NaCl. When you do the experiment yourself, you find that you only produce 3.98 g NaCl. What is the percent yield?Explanation / Answer
9. a. Mg(NO3)2 one mole (148.3 g) contains 1 mole Mg (24 g/mole), 1 mole N2 (28 g/mole) and 3 moles O2 (32 g/mole)
Percentage compositions:
Mg: 24/148.3 X100 = 16.2%
N2: 28/148.3 X100 = 18.9%
O2: 96/148.3 X100 = 64.7%
b) Al2S3 one mole (150.2 g) contains 2 moles of Al (27 g/mole) and 3 moles of S (32 g/mole) atoms
Percentage compositions:
Al: 54/150.2 X100 = 36%
S: 96/150.2 X100 = 64%
10. stock HCl concentration = 15.8 M
diluted HCl solution concentration = 0.75 M
its volume = 3.5 L
volume of 15.8 M HCl needed to make 0.75 M HCl solution of .5 L
V1 = M2V2/M1 = 0.75 X 3.5/ 15.8 = 0.166 L (or) 166 mL
11) By the stoichiometric equation given, 5 moles of Fe(II) needs 1 mole of Mn(VII) for its titration.
the given wt. of FeSO4 = 3.36 g
moles of FeSO4 = 3.36/151.9 (1 mole) = 0.022 mole
the required moles of KMnO4 = 0.022/5 = 0.0044 moles
volume of 0.25 M KMnO4 required = moles/M = 0.0044/0.25 = 0.0176 L (or) 17.6 mL
12) 0.25 M Ba(NO3)2 contains 0.25 M Ba+2, 0.5 M NO3- ionic concentrations;
0.5 M CeCl contains 0.5 M Cs+ and 0.5 M Cl-;
0.15 M Al(OH)3 contains 0.15 M Al+3 and 0.45 M 0f OH- ions.
13. Percnt yield = produced amount/stoichiometric amount X100 = 3.98/5.9 X100 = 67.4%
Best wishes.
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