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alevelhelpxoxo
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The mineral cerussite consists of lead II carbonate, PbCO3. A particular lead ore consisted only of cerissute and inert rocky material. 1.00 g of crushed ore was added to 25cm3 of 1 moldm-3 nitric acid (an excess) in a beaker. A solution is formed and is made up to 250cm3 with water. A 25cm3 sample of this was titrated with 0.100 moldm-3 NaOH. 20.5 cm3 was required to produce a neutral solution.
PbCO3 + 2HNO3 = Pb(NO3)2 + CO2 + H2O
HNO3 + NaOH = NaNO3 + H2O
a) calculate the number of moles of the nitric acid present in the original 25cm3 of 1 moldm-3 acid
b) calculate the number of moles of nitric acid left after the reaction with ore
c) calculate the no. of moles of lead II carbonate in 1g of the ore
d) calculate the mass of lead which can be obtained from 1 tonne of the ore , assuming 100% extraction was possible.
PbCO3 + 2HNO3 = Pb(NO3)2 + CO2 + H2O
HNO3 + NaOH = NaNO3 + H2O
a) calculate the number of moles of the nitric acid present in the original 25cm3 of 1 moldm-3 acid
b) calculate the number of moles of nitric acid left after the reaction with ore
c) calculate the no. of moles of lead II carbonate in 1g of the ore
d) calculate the mass of lead which can be obtained from 1 tonne of the ore , assuming 100% extraction was possible.
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charco
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#2
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#2
(Original post by gcsehelpxoxo)
The mineral cerussite consists of lead II carbonate, PbCO3. A particular lead ore consisted only of cerissute and inert rocky material. 1.00 g of crushed ore was added to 25cm3 of 1 moldm-3 nitric acid (an excess) in a beaker. A solution is formed and is made up to 250cm3 with water. A 25cm3 sample of this was titrated with 0.100 moldm-3 NaOH. 20.5 cm3 was required to produce a neutral solution.
PbCO3 + 2HNO3 = Pb(NO3)2 + CO2 + H2O
HNO3 + NaOH = NaNO3 + H2O
a) calculate the number of moles of the nitric acid present in the original 25cm3 of 1 moldm-3 acid
b) calculate the number of moles of nitric acid left after the reaction with ore
c) calculate the no. of moles of lead II carbonate in 1g of the ore
d) calculate the mass of lead which can be obtained from 1 tonne of the ore , assuming 100% extraction was possible.
The mineral cerussite consists of lead II carbonate, PbCO3. A particular lead ore consisted only of cerissute and inert rocky material. 1.00 g of crushed ore was added to 25cm3 of 1 moldm-3 nitric acid (an excess) in a beaker. A solution is formed and is made up to 250cm3 with water. A 25cm3 sample of this was titrated with 0.100 moldm-3 NaOH. 20.5 cm3 was required to produce a neutral solution.
PbCO3 + 2HNO3 = Pb(NO3)2 + CO2 + H2O
HNO3 + NaOH = NaNO3 + H2O
a) calculate the number of moles of the nitric acid present in the original 25cm3 of 1 moldm-3 acid
b) calculate the number of moles of nitric acid left after the reaction with ore
c) calculate the no. of moles of lead II carbonate in 1g of the ore
d) calculate the mass of lead which can be obtained from 1 tonne of the ore , assuming 100% extraction was possible.
What is the problem exactly?
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virtualcounselor
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a) moles = conc. × volume (in dm^3)
moles = 1.00 mol dm^-3 × (25cm^3 ÷ 1000)
moles = 0.025 mol of nitric acid
________________________________________________
b) as you can see in the equation, you initially start with 2 moles of nitric acid and then you're left with 1 mole of nitric acid on the product side so...
I think you just divide the number of moles in a) by 2:
no. of moles left = 0.025 mol ÷ 2
moles = 0.0125 moles of nitric acid left
________________________________________________
c) moles = mass in grams ÷ Mr (molar mass)
moles = 1.00 gram ÷ (207 plus 12 plus (16 × 3))
moles = 1.00 gram ÷ 267 gmol^-1 (this is the unit of molar mass)
moles = 0.00375 mol of PbCO3
________________________________________________
and... I don't know how to do question d)
*** sorry if my answer or working out is incorrect, I just wanted to help you out
***
moles = 1.00 mol dm^-3 × (25cm^3 ÷ 1000)
moles = 0.025 mol of nitric acid
________________________________________________
b) as you can see in the equation, you initially start with 2 moles of nitric acid and then you're left with 1 mole of nitric acid on the product side so...
I think you just divide the number of moles in a) by 2:
no. of moles left = 0.025 mol ÷ 2
moles = 0.0125 moles of nitric acid left
________________________________________________
c) moles = mass in grams ÷ Mr (molar mass)
moles = 1.00 gram ÷ (207 plus 12 plus (16 × 3))
moles = 1.00 gram ÷ 267 gmol^-1 (this is the unit of molar mass)
moles = 0.00375 mol of PbCO3
________________________________________________
and... I don't know how to do question d)
*** sorry if my answer or working out is incorrect, I just wanted to help you out
***
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