Question

8 The Following reaction is classified as: 2HCl(aq) + Ba(OH)2(aq) → BaCl(aq) + 2 H2000 a) oxidation-reduction reaction b) precipitation reaction c) single replacement d) acid-base neutralization reaction What is the volume of 0.1220 M HINO, solution required to completely neutralize 17.50 ml of 0.1211 M NaOH ? Reaction is: HNO, (aq) + NaOH(aq) - NaNO, (ng) + H. (A) a) 17.37 ml. b) 16.31 ml. c) 15.39 mL d) 17.84 mL 10. Which one of the following reactions is the reduction reaction? a) Cu" (aq) + 2e + Cu() b) Cr() - Ce(aq) + 3 c) Cu → Cut të d) Fe Felt +2 11. Using the following portion of the activity series of elements Ca(s)-Cat(aq) + 2e- (strongly reducing) Al(s)-AP (aq) + 3e Cu(s) - Cu(aq) + 2e (weakly reducing) Determine which one of the reactions will occur. a) Cu(s) + Ca(aq) b) Al(s) + Cu?+ (aq) c) Al(s) + Ca?"( aq d ) Cu(s) +APP(aq) In the following combustion reaction of methane gas (CH) with oxygen is given How many (g) of H2O is produced when 0.533 moles of CH, reacts with O,? 12. a) 15.23g b) 9.60 c) 19.21 g d) 17.21 g 13 How many grams (R) of NaCl are needed to prepare 250.0 ml of 0.365 M of NaCl Solution! a) 5.34 g b) 10 25 g c)3.65 g d) 7.218

Answer 1

Answer:-

(8)-

As we know that HCl_(aq) is an acid and Ba(OH)_2(aq) is a base therefore given reaction is acid-base neutralization reaction in which salt BaCl_2(aq) and water (H₂O_(l)) are formed.

2HCl_(aq) + Ba(OH)_2(aq)  $\rightarrow$ BaCl_2(aq) + 2H₂O_(l)

Therefore correct option is 'd' i.e acid-base neutralization reaction (i.e the answer).

(9)-

Given:-

volume of NaOH (V₁) = 17.50 mL

molarity of NaOH (M₁) = 0.1211 M

volume of HNO₃ (V₂) = 0.1220 M

molarity of HNO₃ (M₂) = ?

According to the formula

M₁V₁ = M₂V₂

So

0.1211 M $\times$ 17.50 mL =   0.1220 M $\times$ V₂

volume of HNO₃ (V₂) = 0.1211 M $\times$ 17.50 mL / 0.1220 M

volume of HNO₃ (V₂) = 2.11925‬ mL / 0.1220

volume of HNO₃ (V₂) = 17.37 mL

Therefore correct option is 'a' i.e 17.37 mL (i.e the answer).

(10)-

As we know that Reduction reaction in which an element or ion has tendency to accept the electrons to get reduced. In this reaction, the oxidation state (oxidation no.) of  element or ion is decreased whereas oxidation reaction in which an element or ion has tendency to donate the electrons to get oxidized. In this reaction, the oxidation state (oxidation no.) of  element or ion is increased.

So

(a) -

Cu²⁺_(aq) + 2e- $\rightarrow$ Cu(s) (reduction)

+2 0

(b) -

Cr_(s) $\rightarrow$ Cr³⁺_(aq) + 3e- (oxidation)

0 +3

(c) -

Cu⁺_(aq)$\rightarrow$ Cu²⁺_(aq) + e- (oxidation)

+1 +2

(d) -

Fe_(s)$\rightarrow$ Fe²⁺_(aq) + e- (oxidation)

0 +2

Therefore correct option is 'a' i.e Cu²⁺_(aq) + 2e- $\rightarrow$ Cu(s)   (i.e the answer).

(11)-

As we know that Reduction reaction in which an element or ion has tendency to accept the electrons to get reduced. In this reaction, the oxidation state (oxidation no.) of  element or ion is decreased therefore this reduced element or ion helps other element or ion to get oxidized and hence such element or ion which reduced are called as oxidizing agent. whereas oxidation reaction in which an element or ion has tendency to donate the electrons to get oxidized. In this reaction, the oxidation state (oxidation no.) of  element or ion is increased. therefore this oxidized element or ion helps other element or ion to get reduced and hence such element or ion which oxidized are called as reducing agent.

So

Ca_(s)$\rightarrow$ Ca²⁺_(aq) + 2e- (oxidation) i.e strong reducing agent

Al_(s)$\rightarrow$ Al³⁺_(aq) + 3e- (oxidation) i.e moderate reducing agent

Cu_(s) $\rightarrow$ Cu²⁺_(aq) + 2e- (oxidation) i.e weak reducing agent

So order of reducing agent is as follows:-

Ca(s)  > Al(s) >   Cu(s)

Also we know that strong reducing agent oxidized weak reducing agent therefore

(b)- Al_(s) + Cu²⁺_(aq) (reaction will occur)

strong reducing agent   weak reducing agent

In above Al_(s) is strong reducing agent as compared to the Cu²⁺_(aq) (weak reducing agent)

So correct option is 'b' i.e Al_(s) +  Cu²⁺_(aq) (i.e the answer).

(12)-

As we know that the combustion reaction of methane (CH₄) with oxygen (O₂) is as follows:-

CH_4(g) + 2O_2(g)    $\rightarrow$ CO_2(g) + 2H₂O _(l)

1 mol 2 mol 1 mol 2 mol

16 g 2 $\times$ 32 g 44 g    2 $\times$ 18 g

16 g 64 g 44 g 36 g

As mentioned above

1 mol of CH₄ reacts with O₂ to produced = 36 g H₂O

then

0.533 mol of CH₄ reacts with O₂ to produced = 0.533 $\times$ 36 g H₂O

0.533 mol of CH₄ reacts with O₂ to produced = 19.19 g H₂O which equivalent to the 19.21 g H₂O

So correct option is 'c' i.e 19.21 g (i.e the answer).

(13)-

Given:-

volume of NaCl solution (V) = 250 mL = 250 / 1000 L = 0.25 L

molarity of NaCl (M) = 0.365 M = 0.365 mol /L

wt. of NaCl (w) = ?

As we know that

molar mass of NaCl (m) = molar mass of Na + molar mass of Cl

molar mass of NaCl (m) = 23 + 35.5

molar mass of NaCl (m) = 58.5 g / mol

Also we know that

molarity of compound (M) = wt. of compound (w) / molar mass of compound (m) $\times$ volume of compound solution (V) in liter

therefore

wt. of compound (w) =  molarity of compound (M) $\times$ molar mass of compound (m) $\times$ volume of compound solution (V) in liter

So

wt. of NaCl (w) =  molarity of NaCl (M) $\times$ molar mass of NaCl (m) $\times$ volume of NaCl solution (V) in liter

wt. of NaCl (w) =  0.365 mol /L $\times$ 58.5 g / mol  $\times$ 0.25 L

wt. of NaCl (w) =  0.365 mol /L $\times$ 58.5 g / mol  $\times$ 0.25 L

wt. of NaCl (w) = 5.34 g

So correct option is 'a' i.e 5.34 g (i.e the answer).