Question

24. You have 135.8 mL of a 2.50 M solution of Na Cros(am). You also have 125 mL of a 2.50 M solution of AgNO,Cao). Calculate the concentration of Nat after the two solutions are mixed together a) 0.00 M b) 1.30 M c) 2.60 M d) 5.00 M e) 0.679 M The oxidation state of chlorine in CIO;- is: b) c) +5 5 26. In the reaction Zn + H2SO4 → ZnSO4 + Hy, which, if any, element is oxidized? a) zinc b) hydrogen c) sulfur d) oxygen e) none of these 27. Balance the following oxidation-reduction reaction using the oxidation number method: Al+Br, → AB + Br In the balanced equation, the coefficient of Br is b) c) 3 4 e) none of these 28. Determine the mass of CO, produced when 75.5 g of CaQ is reacted with 50.0 g of C according to the unbalanced equation CaO+C → CaC + CO2 a 29.6 g CO, b. 119 g CO2 C. 59.3 g CO2 d. 36.6 g CO. e. none of these 29. Consider the following reaction: 2A + B 3C+D 3.35 mol A and 2.00 mol B react to form 4.00 mol C. What is the percent yield of this reaction? a 20.4%

Answer 1

24)

Na2CrO4 is a salt that undergoes complete ionisation

$Na_2CrO_4\rightarrow 2Na^+ \, \, \, \, \, +\, \, \, \, \, CrO_4^^^2^^^-$

according to this equation,

1 Mole  Na2CrO4  gives 2 moles of Na+

The number of moles of  Na+ in 1L of 2.5M Na2CrO4 = 2*2.5 moles

Hence, The number of moles of  Na+ in 135.8mL of 2.5M Na2CrO4 = 2*2.5*0.1358 moles = 0.68moles

When we 135.8mL of 135.8mL of 2.5M Na2CrO4 and 125mL of 2.5M AgNO3 , we have:

Total volume of solution = 135.8+125 = 260.8mL = 0.2608L

Since Na+ comes only from 135.8mL of 2.5M Na2CrO4 , the number of moles of  Na+ in solution = 0.68 moles

concentration of of  Na+ in solution = 0.68 / 0.2608 = 2.60 M

Hence answer for Question-24 is c) 2.60 M

25)

Let oxidation state of Cl = x

oxidation state of each O = -2

Total charge of molecule = -1

Hence we can write ;

$x+3 \times (-2)=-1$

OR

$x-6=-1$

OR

$x=+5$

Hence oxidation state of Cl in ClO3-= +5

Hence answer for Question-25 is b) +5

26)

Here, on the left-hand side of the reaction, Oxidation State of Zn = 0

on the right-hand side of the reaction, Oxidation State of Zn = +2

Since the Oxidation State of Zn has increased it is oxidised.

For all other elements, the Oxidation state either decreases( in case of H , it decrease from +1 to 0) or remains the same ( in case of S and O ,it remains the same) . Hence no other element is oxidised.

Hence answer for Question-26 is a) zinc

27)

The balanced equation is written as

$2\; Al\: +\:3 \:Br_2 \rightarrow 2\:Al^3^+ \:+\:6 \:Br^-$

Hence it is clear that the coefficient of Br- is 6

Hence answer for Question-27 is d) 6