Which of the statements about the reaction below are incorrect ?
$2 P b O (s) + C (s) \to 2 P b (s) + C O_{2} (g)$
(a) Lead is getting reduced.
(b) Carbon dioxide is getting oxidised.
(c) Carbon is getting oxidised.
(d) Lead oxide is getting reduced

(a) and (b)

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(a) and (c)

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(a), (b) and (c)

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All

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Solution

The correct option is C (a) and (b)
$2 P b O (s) + C (s) \to 2 P b (s) + C O_{2} (g)$
+2 0 0 +4
$P b O$ reduces to $P b$ and $C$ oxidises into $C O_{2}$ .

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Similar questions

Q. Which of the given statement(s) is/are correct about the reaction mentioned below?

2 PbO (s) + C (s) \to 2 Pb (s) + {CO}_{2} (s)

Q. Which of the statements about the reaction below are incorrect?

2 P b O (s) + C (s) \to 2 P b (s) + C O_{2} (g)

(a) Lead is getting reduced
(b) Carbon dioxide is getting oxidized
(c) Carbon is getting oxidized
(d) Lead oxide is getting reduced

Which of the given statements are correct about the reaction mentioned below?
$2 P b O_{(s)} + C_{(s)} \to 2 P b_{(s)} + C O_{2 (s)}$
i) Lead is getting reduced
ii) Carbon dioxide is getting oxidised
iii) Carbon is getting oxidised
iv) Lead oxide is getting reduced

Q. Question 1
Which of the statements about the reaction below are incorrect?

2 PbO(s) + C(s) \to 2 Pb (s) + {CO}_{2} (g)

(a) Lead is getting reduced.
(b) Carbon dioxide is getting oxidized.
(c) Carbon is getting oxidized.
(d) Lead oxide is getting reduced.

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Which of the statements about the reaction below are incorrect ?2PbO(s)+C(s)→2Pb(s)+CO2(g)(a) Lead is getting reduced.(b) Carbon dioxide is getting oxidised.(c) Carbon is getting oxidised.(d) Lead oxide is getting reduced

The correct option is C (a) and (b)2PbO(s)+C(s)→2Pb(s)+CO2(g) +2 0 0 +4PbO reduces to Pb and C oxidises into CO2.

Which of the statements about the reaction below are incorrect ?
$2 P b O (s) + C (s) \to 2 P b (s) + C O_{2} (g)$
(a) Lead is getting reduced.
(b) Carbon dioxide is getting oxidised.
(c) Carbon is getting oxidised.
(d) Lead oxide is getting reduced

The correct option is C (a) and (b)
$2 P b O (s) + C (s) \to 2 P b (s) + C O_{2} (g)$
+2 0 0 +4
$P b O$ reduces to $P b$ and $C$ oxidises into $C O_{2}$ .