hayz Posted March 15, 2006 Share Posted March 15, 2006 hey guys and guls, i was just wondering if you can show me how to work this question out? consider the following reaction: H2S(g) + 2 NaOH(aq) -> Na2S(aq) + 2 H2O(l) How many grams of Na2S are formed if 2.05 g of H2s is bubbled into a solution containing 1.84 g of NaOH, assuming the limiting reagent is completely consumed? Be great if you can help, im having problems working out these questions. Hayz Link to comment Share on other sites More sharing options...
RyanJ Posted March 15, 2006 Share Posted March 15, 2006 Firstly work out the Mr for each of the compounds then read this - its an example reaction and it should help you understand how it works http://en.wikipedia.org/wiki/Mole_%28unit%29#Example_calculation ...and welcome too SFN! Cheers, Ryan Jones Link to comment Share on other sites More sharing options...
Dak Posted March 15, 2006 Share Posted March 15, 2006 What's Mr? Link to comment Share on other sites More sharing options...
YT2095 Posted March 15, 2006 Share Posted March 15, 2006 Moles. Link to comment Share on other sites More sharing options...
hayz Posted March 15, 2006 Author Share Posted March 15, 2006 cheers Ryan your example helped but wasnt quite the same so i am still stuck with it but il keep trying to see wot i get Link to comment Share on other sites More sharing options...
dttom Posted March 15, 2006 Share Posted March 15, 2006 H2S + 2NaOH --> 2H2O + Na2S mole ratio of H2S:NaOH is 1:2; 2.05g H2S: 2.05/(2+32)mol=0.06mol; 1.84g NaOH: 1.84/40mol=0.046mol; so 0.023mol of H2S is reacted, mole ratio of H2S to Na2S is 1:1; 0.023mol of Na2S is formed, =0.023*(23*2+32)=1.794g Link to comment Share on other sites More sharing options...
Dak Posted March 15, 2006 Share Posted March 15, 2006 Moles. Cheers -- i've never seen it written like that. ==================== If you can't make out what dttom did: 1st, he worked out how many mols were present, by dividing the weight of the substance by the weight-per-mol* of that substance. Then he worked out the limiting reagent: theres less mols of NaOH than of H2S, and so not all of the H2S will be reacted ('cos there isn't enough NaOH for it to react with), so NaOH is the limiting reagent. That means that all 0.046mols of NaOH will be reacted; as two mols of NaOH react with one mol of H2S, this means that only 0.023mols of H2S will react. dttom then noted, from the equasion, that each mol of H2S will yield a mol of Na2S; hence the reaction will yield 0.023mols Na2S. You then simply convert mols of Na2S into grams of Na2S by timesing the number of mols (0.023) by the weight-per-mol* of Na2S. ----------- *look it up in a book, or look up the weight-per-atom of the constituent elements and add them up, eg Na2S = 2*(weght-per-atom of sodium)+(weight-per-atom of sulphur). Link to comment Share on other sites More sharing options...
hayz Posted March 16, 2006 Author Share Posted March 16, 2006 thanx so much guys, you have been a great help. i understand it now. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now