ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: Starlight on June 16, 2012, 08:58:12 pm
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Chlorine can be prepared via: Mn02 (s) + 4 Hcl (aq) ---> Cl2(g) + MnCl2(aq) + 2 H20
How many moles of Cl2 can be prepared from 0.75 mol Mn02 and 2 mol Hcl?
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0.5 mole im pretty sure
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Lets see if I remember any chem. *b^3 dusts off brain*
Mole Ratio b/w MnO2 and HCl is 1:4 for the equation, but we only have (1/0.75)*2=2.67 mol of HCl for every 1 mol of MnO2, so that makes HCl our limiting reagent.
So the mol ratio b/w HCl and Cl2 is 4:1, so 4 mols of HCl makes 1 mol of CL2, we have 2 mols of HCl, so 2/4=x/1 ->x=0.5 mol as ligands said above.
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Stick, you forgot one is in excess
edit: he removed his post -.-
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Lets see if I remember any chem. *b^3 dusts off brain*
Mole Ratio b/w MnO2 and HCl is 1:4 for the equation, but we only have (1/0.75)*2=2.67 mol of HCl for every 1 mol of MnO2, so that makes HCl our limiting reagent.
So the mol ratio b/w HCl and Cl2 is 4:1, so 4 mols of HCl makes 1 mol of CL2, we have 2 mols of HCl, so 2/4=x/1 ->x=0.5 mol as ligands said above.
i did it a different way lol, i multiplied the mole of MnO2 by 4 so the ratios were the same 3 mol and 2 moles therefore HCL was limiting, is this a good method of going it? or no
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Lets see if I remember any chem. *b^3 dusts off brain*
Mole Ratio b/w MnO2 and HCl is 1:4 for the equation, but we only have (1/0.75)*2=2.67 mol of HCl for every 1 mol of MnO2, so that makes HCl our limiting reagent.
So the mol ratio b/w HCl and Cl2 is 4:1, so 4 mols of HCl makes 1 mol of CL2, we have 2 mols of HCl, so 2/4=x/1 ->x=0.5 mol as ligands said above.
i did it a different way lol, i multiplied the mole of MnO2 by 4 so the ratios were the same 3 mol and 2 moles therefore HCL was limiting, is this a good method of going it? or no
That's what I do
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Lets see if I remember any chem. *b^3 dusts off brain*
Mole Ratio b/w MnO2 and HCl is 1:4 for the equation, but we only have (1/0.75)*2=2.67 mol of HCl for every 1 mol of MnO2, so that makes HCl our limiting reagent.
So the mol ratio b/w HCl and Cl2 is 4:1, so 4 mols of HCl makes 1 mol of CL2, we have 2 mols of HCl, so 2/4=x/1 ->x=0.5 mol as ligands said above.
i did it a different way lol, i multiplied the mole of MnO2 by 4 so the ratios were the same 3 mol and 2 moles therefore HCL was limiting, is this a good method of going it? or no
Either should work, doesn't really matter, although I have forgotten a lot of chem, so as to the best method......? (although they are nearly the same thing, just backwards of each other to arrive at the same place).
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Stick, you forgot one is in excess
edit: he removed his post -.-
Removed it since I was embarrassed I don't have the knowledge to do it as a Year 11 student. lol