Find the mass of lead formed by the reduction of 342.5 g of red lead( $P b_{3} O_{4}$ ) in a current of hydrogen.

210.5 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

310.5 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

250 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

377.79 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is
B

310.5 g

$P b_{3} O_{4}$ + $4 H_{2}$ $\to$ 3Pb + $4 H_{2} O$
(207 $\times$ 3+16 $\times$ 4) 3 $\times$ 207
685 g 621g

685 g of red lead yields 621 g of lead.
1 g of red lead yields $\frac{621}{685}$ g of lead.

342.5 g of red lead = $\frac{621 \times 342.5}{685}$

= 310.50 g of lead.

Suggest Corrections

Similar questions

Q. Manganese (IV) oxide, lead (IV) oxide and red lead

[P b_{3} O_{4}]

react with conc.

H C I

liberating chlorine.Write the equation for the reaction of conc.

H C l

with

P b_{3} O_{4}

Manganese (IV) oxide, lead (IV) oxide and red lead (Pb₃O₄) react with concentrated hydrochloric acid to liberate chlorine.
(a) Name the common property exhibited by these metal oxides.
(b) Write the equation for the reaction of concentrated hydrochloric acid with Pb₃O₄.

Q. Two oxides of lead were separately reduced to metallic lead by heating in a current of, hydrogen and the following data obtained :
i) Mass of yellow oxide taken

= 3.45 g

Loss in mass during reduction

= 0.24 g

ii) Mass of brown oxide taken

= 1.227 g

, Loss in mass during reduction

= 0.16 g

Show that the above data illustrate the Law of Multiple Proportions.

Q. Manganese (IV) oxide, lead (IV) oxide and red lead

[P b_{3} O_{4}]

react with conc.

H C I

liberating chlorine. What is the common property being shown by these metal oxides.

The reaction between red lead and hydrochloric acid is given below:
$P b_{3} O_{4} + 8 H C L I \to 3 P b C l_{2} + 4 H_{2} O + C l_{2}$
calculate : (a) the amss fo lead chloride formed by the action of 6.85 g of red lead, (b) the mass fo chlorine and (c) the volume of chlorine evolved at STP.

Join BYJU'S Learning Program

Find the mass of lead formed by the reduction of 342.5 g of red lead(Pb3O4) in a current of hydrogen.

The correct option is B 310.5 g Pb3O4 + 4H2 → 3Pb + 4H2O (207×3+16×4) 3×207 685 g 621g 685 g of red lead yields 621 g of lead.1 g of red lead yields 621685 g of lead. 342.5 g of red lead = 621×342.5685 = 310.50 g of lead.

Find the mass of lead formed by the reduction of 342.5 g of red lead( $P b_{3} O_{4}$ ) in a current of hydrogen.