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### Topic: Stoichiometry Question  (Read 10087 times)

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#### positiveion

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##### Stoichiometry Question
« on: January 06, 2010, 03:41:52 AM »
I don't know how to do this question mainly because I don't understand the last equation with all its dashes and such. I don't really know where to begin or what to do.

#### positiveion

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##### Re: Stoichiometry Question
« Reply #1 on: January 06, 2010, 04:12:09 AM »
I have another few questions:

The value of x in Fe(NH4)2(SO4)2.xH2O can be found by determining the amount in moles of sulfate in the compound.

A 0.982 sample was dissolved in water and excess BaCL2 was added. The precipitate of BaSO4 was seperated and dried and found to weigh 1.17g.

Calculate the amount in moles of sulfate in the 0.982g sample of Fe(NH4)2(SO4)2.xH2O

What I have been doing is"
a). no. moles of the entire thing: 0.982 / 284.07 = 0.003456
b). (192.12/284.07) = 0.6763121766 [192.12 is the mass of (SO4)2
c) (a) * (b) =0.002 which i thought was the answer but the answer is actually 0.00501

#### AWK

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##### Re: Stoichiometry Question
« Reply #2 on: January 06, 2010, 04:13:18 AM »
I don't know how to do this question mainly because I don't understand the last equation with all its dashes and such. I don't really know where to begin or what to do.

Show your attempt. This is an elementary stoichiometry problem.
AWK

#### AWK

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##### Re: Stoichiometry Question
« Reply #3 on: January 06, 2010, 04:18:03 AM »
I have another few questions:

The value of x in Fe(NH4)2(SO4)2.xH2O can be found by determining the amount in moles of sulfate in the compound.

A 0.982 sample was dissolved in water and excess BaCL2 was added. The precipitate of BaSO4 was seperated and dried and found to weigh 1.17g.

Calculate the amount in moles of sulfate in the 0.982g sample of Fe(NH4)2(SO4)2.xH2O

What I have been doing is"
a). no. moles of the entire thing: 0.982 / 284.07 = 0.003456
b). (192.12/284.07) = 0.6763121766 [192.12 is the mass of (SO4)2
c) (a) * (b) =0.002 which i thought was the answer but the answer is actually 0.00501

Start from a balanced reaction
AWK

#### positiveion

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##### Re: Stoichiometry Question
« Reply #4 on: January 06, 2010, 04:26:54 AM »
I don't know how to do this question mainly because I don't understand the last equation with all its dashes and such. I don't really know where to begin or what to do.

Show your attempt. This is an elementary stoichiometry problem.

I don't know what to do because I don't know what the DASHES mean.

#### positiveion

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##### Re: Stoichiometry Question
« Reply #5 on: January 06, 2010, 04:43:43 AM »
I have another few questions:

The value of x in Fe(NH4)2(SO4)2.xH2O can be found by determining the amount in moles of sulfate in the compound.

A 0.982 sample was dissolved in water and excess BaCL2 was added. The precipitate of BaSO4 was seperated and dried and found to weigh 1.17g.

Calculate the amount in moles of sulfate in the 0.982g sample of Fe(NH4)2(SO4)2.xH2O

What I have been doing is"
a). no. moles of the entire thing: 0.982 / 284.07 = 0.003456
b). (192.12/284.07) = 0.6763121766 [192.12 is the mass of (SO4)2
c) (a) * (b) =0.002 which i thought was the answer but the answer is actually 0.00501

Start from a balanced reaction

Fe(NH4)2(SO4)2 + BaCl2 --> BaSO4 + Fe(NH4)Cl2

I'm not sure how to balance that because when I add a 2 in front of BaSO4 then I need to add one in front of BaCl2 and then I need to add one in front of Fe(NH4)2Cl2 and then to Fe(NH4)2(SO4)2 -- then i need to change BaSO4 again and its an endless cycle.

#### AWK

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##### Re: Stoichiometry Question
« Reply #6 on: January 06, 2010, 08:07:00 AM »
Quote
Fe(NH4)2(SO4)2 + BaCl2 --> BaSO4 + Fe(NH4)Cl2

Fe(NH4)2(SO4)2 + 2BaCl2 =  2BaSO4 + FeCl2 + 2NH4Cl
AWK

#### cliverlong

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##### Re: Stoichiometry Question
« Reply #7 on: January 06, 2010, 08:55:26 AM »

I don't know what to do because I don't know what the DASHES mean.
Look in the first few paragraphs of the following

http://www.chemguide.co.uk/organicprops/alkenes/polymerisation.html