If 10.0mL. of phosphoric acid solution required 23.5mL of 0.200M sodium hydroxid
ID: 983450 • Letter: I
Question
If 10.0mL. of phosphoric acid solution required 23.5mL of 0.200M sodium hydroxide solution for neutralization to a phenolphthalein end pointt, calculate the molarity of the phosphoric acid. H3PO4(aq) + 3NaOH(aq) rightarrow Na3PO4(aq) + 3H20(1) a. 0.157 M b. 0.470 M c. 1.41 M f. 1.17 M e. none of these 11. What mass of NaOH is needed to precipitate the Cd^- ions from 35.0 mL of 0.500 M Cd(NO_3)_2 solutic Cd(NO3)_2 (aq) + 2 NaOh_(aq) rightarrow 2 NaNO3_(aq) + Cd(OH)_2 (s) a. 0.0175 g b. 1.40 g c. 0.0350 g d. 0.700 g e. none of theseExplanation / Answer
10) consider the given reaction
H3P04 + 3 NaOH ---> Na3P04 + 3 H20
we can see that
moles of NaOH = 3 x moles of H3P04
also
moles = molairty x volume (L)
so
M x V of NaOH = 3 x M x V of H3P04
so
0.2 x 23.5 = 3 x M x 10
M = 0.157
so
a) 0.157 M
11) we can see that
moles of NaoH = 2 x moles of Cd(N03)2
so
moles of NaOH = 2 x M x V of Cd(N03)2
moles of NaoH = 2 x 0.5 x 35 x 10-3
moles of NaOH = 35 x 10-3
now
mass = moles x molar mass
so
mass of NaOH = 35 x 10-3 x 40
mass of NaOH = 1.4 g
so
b) 1.40 g
12)
moles of Mg(OH)2 = 2.87 / 58.3197
moles of Mg(OH)2 = 0.0492
now
the reaction is =
Mg(OH)2 + 2HCl ---> MgCl2 + 2H20
so
moles of HCl = 2 x moles of Mg(OH)2
moles of HCl = 2 x 0.0492
moles of HCl = 0.0984
now
volume = moles / molarity
so
volume of HCl = 0.0984 / 0.128
volume of HCl = 0.769 L
volume of HCl = 769 ml
so
d) 769 ml
13) the reaction is
HCl04 + NaOH ---> NaCl04 + H20
so
moles of HCl04 = moles of NaoH
M x V of HCl04 = M x V of NaoH
so
0.115 x V = 0.0875 x 50
V = 38.04 ml
so
C) 38 ml
14)
moles of CaCl2 = 785 x 10-3 / 111 = 7.072 x 10-3
now
moles of Cl- = 2 x 7.072 = 14.144 x 10-3
now
the reaction is
Ag+ + Cl- --> AgCl
so
moles of Ag+ = moles of Cl-
moles of Ag+ = 14.144 x 10-3
now
moles of AgN03 = moles of Ag+ = 14.144 x 10-3
so
molarity = moles / volume (L)
so
molarity = 14.144 x 10-3 / 25.8 x 10-3
molarity = 0.548
so
a) 0.548 M
15)
the reaction is
2 Fe + 3Br2 ---> 2 FeBr3
Fe ---> Fe+2 so Fe is oxidized , so reducing agent
Br2 --> Br- so Br is reduced , so oxidizing agent
so
the answer is option B
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