Suki123z
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9. Lead (IV) oxide dissolves in concentrated hydrochloric acid according to the following equation:
PbO2(s) + 4HCl(aq) à PbCl2(s) + Cl2(g) + 2H2O(l)

Starting with 37.2 g of lead (IV) oxide, calculate:

a) the volume of 12 moldm-3 HCl needed to completely dissolve it

b) the mass of PbCl2 produced

c) the volume of chlorine produced at 298 K and 100 kPa.

I am stuck on this question could someone please help me, thanks.
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charco
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(Original post by Suki123z)
9. Lead (IV) oxide dissolves in concentrated hydrochloric acid according to the following equation:
PbO2(s) + 4HCl(aq) à PbCl2(s) + Cl2(g) + 2H2O(l)

Starting with 37.2 g of lead (IV) oxide, calculate:

a) the volume of 12 moldm-3 HCl needed to completely dissolve it

b) the mass of PbCl2 produced

c) the volume of chlorine produced at 298 K and 100 kPa.

I am stuck on this question could someone please help me, thanks.
Which part exactly are you stuck on?
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Crunch_chemistry
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(Original post by Suki123z)
9. Lead (IV) oxide dissolves in concentrated hydrochloric acid according to the following equation:
PbO2(s) + 4HCl(aq) à PbCl2(s) + Cl2(g) + 2H2O(l)

Starting with 37.2 g of lead (IV) oxide, calculate:

a) the volume of 12 moldm-3 HCl needed to completely dissolve it

b) the mass of PbCl2 produced

c) the volume of chlorine produced at 298 K and 100 kPa.

I am stuck on this question could someone please help me, thanks.
a) work out mol of lead oxide (mol = mass / Mr), determine ratio of lead oxide : HCl from equation, work out volume (in dm3) by rearranging mol = c x v
b) as above
c) 298K and 100kPa is room temp and pressure. 1 mol of any gas occupies 24.0 dm3 under these conditions. Work out mol of chlorine following method above and then scale up / down from 24.0.
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