[ryan1]

A fertiliser contains ammonium sulphate. a sample of 0.5g of fertiliser was warmed wit hsodium hydroxide solution. the ammonia evolved waas absorbed in 100cm3 of 0.1 moldm3 hydrochloric acid. the excess of hydrochloric acid required 55.9 cm3 of 0.1 mol dm3 sodium hydroxide for neutralisatioin. calculate the percentage of ammonium sulphate in the sample.



NaOH moles: (55.9/1000)x0.1=5.59X10-3moles

moels of HCl:(100/1000)x0.1=0.01 moles

1:2 reaction therefore 5.59x1--3/2 = 2.795x10-3 moles of NH42SO4


now im not sure what to do....

[sjb]

A fertiliser contains ammonium sulphate. a sample of 0.5g of fertiliser was warmed wit hsodium hydroxide solution. the ammonia evolved waas absorbed in 100cm3 of 0.1 moldm3 hydrochloric acid. the excess of hydrochloric acid required 55.9 cm3 of 0.1 mol dm3 sodium hydroxide for neutralisatioin. calculate the percentage of ammonium sulphate in the sample.



NaOH moles: (55.9/1000)x0.1=5.59X10-3moles

moels of HCl:(100/1000)x0.1=0.01 moles

So how much HCl reacted with the ammonia?

1:2 reaction therefore 5.59x1--3/2 = 2.795x10-3 moles of NH42SO4


now im not sure what to do....

Not quite sure what you're doing here. 5.59 x 1 - 3/2 is certainly not 2.795 x 10^-3, unless I'm misreading stuff.

[albert academy]

A fertiliser contains ammoniu sulphate .a sample of of 0.5g of the fertiliser was warmed with sodium hydroxide solution .the ammonia evolved was absorbed in 100cm3 of 0.1moldm3 hydrochloric acid .the exess acid required 55.9g cm3 of 0,1 moldm3 sodium hydroxide for the neutralisation. calculate the percentage of the ammonium sulphate pn the sample
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