Recent content by Kiah Palmer

Ok, so then I would need to use Mol SiCl4 then calculate the yield percent? Or do I need to find out what 92% of the mass of Si would be first, then recalculate mol?

so for that, I would need to find Mol Si and mol SiCl4, which I believe is as follows: Theoretical # mol Si = 1.0x10^6 g / 28.085 g/mol = 3.6x10^4 mol Theoretical # Mol SiCl4 = 2.33x10^6 g / 169.897 g/mol = 1.4x10^4 mol I think this would be correct for the # mol (if that was even needed)...

Conversion: 1.0 t = 1.0x10^6 g impure Si Conversion: 2.33 t = 2.33x10^6 g SiCl4 Atomic Mass Si = 28.085 g/mol Atomic Mass Cl2 = (2 x 35.453) = 70.906 g/mol Mass SiCl4 = 28.085 + (4x35.453) = 169.897 g/molI got all this part down, however, I'm not quite sure where I am supposed to go from there...

Ah! ok, so the answer that I had found when looking afterwards is in fact the correct answer! It actually makes sense now! thanks again! :) I did hand it in the way I Have done it, but at least now I will know where I went wrong. :)

Conversion: 9.65 kg = 9650 g Mass Al = 26.98 g/mol Molar Mass Al2O3 = (2 x 26.98) + ( 3 x 16.00) = 101.96 g/mol Mol Al2O3 = 9650 g / 101.96 g/mol = 94.65 mol Mol Al Required = 94.65 mol x 2 = 189.3 mol Mass 100% Al required = 189.3 mol x 26.98 g/mol = 5.107x103 g Mass 80% Al required =...

great! thanks!

Thanks! I actually also sent it to my teacher as well through messages and she said the same. I had sent it here just in case she didn't have a chance to respond. :)

1a) Step 1: Balance the equation - S8 + O2 -> SO2 = S8 + 8O2 -> 8SO2 Conversion (t to g) 6.00 t = 6x10^6 g Molar mass SO2 = 23 + (2 x 16) = 64 g/mol 6x10^6 g / 64 g/mol = 9.375x10^4 mol SO2 reqired amount Sulphur = 9.375x10^4 mol / 8 mol = 1.172x10^4 mol S8 Molar mass S8 = (8 x 32)...

Thanks! :)

Hello everyone! I am new here and just thought I'd take the time to introduce myself. My name is Kiah, and I live in Ontario, Canada. I am currently studying Chemistry through online classes and will be doing Biology next semester. Glad I found this community, as I am quite sure it will prove...