Chemistry help!please!??

[blairep929]

The following reaction is performed in a lab:3Na2SO4(aq) + 2Al(NO3)3(aq) --> Al2(SO4)3(s) + 6NaNO3(aq)If 191 mL of 1.25 M aluminum nitrate is added to an excess of sodium sulfate, how many grams of aluminium sulfate will be produced?HELP! please! This is an extra credit problem on hw and I would love the ec. 10 pts to quickest best answer! TY

[RONALDEB]

0.191 L x 1.25 M = 0.23875 mol Al(NO3)30.23875 mol --> 0.5 x 0.23875 = 0.1194 mol Al2(SO4)30.1194 mol x 342.16 g/mol = 40.8 g Al2(SO4)3

[blazingdarkness16]

First convert to moles .191L X 1.25M= .23875 mol Al(NO3)for every 2 mol of the aluminimum nitrate formed 1 amuminium sulafate is produce thus.23875 mol AL(NO3)(1 MOL AL2(SO4)3/2 MOL AL(NO3)).119375MOL times molecular weight (342g/mol)=40.83grams will be produced.

[shawn101A8929]

0.191L x 1.25mol/L = 0.23875mol Al(NO3)30.23875mol Al(NO3)3 x 1mol Al2(SO4)3 / 2mol Al(NO3)3 = 0.119375 mol Al2(SO4)30.119375 mol Al2(SO4)3 x 342.15 g/mol = 40.84g Al2(SO4)3

[MerlinsFeline]

191mL x 1.25 M = 238.75 mmoles = 0.239 moles of Al(NO3)3in th reaction2(213grams of Al(NO3)3 ) react with Na2SO4 to give 342 grams of Al2(SO4)3so 0.239 moles of AL(NO3)3 = 50.9 grams will yield 426/342=50.9/x = 40.9 gramProof 40.9 grams of Al2(SO4)3 is 40.9/342 grams = 0.1195 molesAl(NO3)3 gives Al2(SO4)3 oin a 2:1 molar ratiomoles of Al(NO3)3 = 0.239 moles of AL2(SO4)3 = 0.11950.239/0.1195 = 2