2 moles of HCl *(36.5 g of HCl/mol)=moles of HCl
Since moles of HCl=moles of Magnesium,
moles of Mg*(24.3g og Mg/mol)= g go Mg
how many grams of magnesium are required to react with 2.0 moles of hydrochloric acid?
6 answers
Mg + 2HCl ==> MgCl2 + H2
2 mols Mg x (2 mols HCl/1 mol Mg) = 4 mols HCl.
g HCl = 4 mols HCl x molar mass HCl. = 4*36.5 = ?
2 mols Mg x (2 mols HCl/1 mol Mg) = 4 mols HCl.
g HCl = 4 mols HCl x molar mass HCl. = 4*36.5 = ?
Opps,
2 moles of HCl *(36.5 g of HCl/mol)=moles of HCl
Since 2 moles of HCl=moles of Magnesium,
moles of HCl *(1 mole of Magnesium/2 moles of HCl)*(24.3g og Mg/mol)= g go Mg
I apologize
2 moles of HCl *(36.5 g of HCl/mol)=moles of HCl
Since 2 moles of HCl=moles of Magnesium,
moles of HCl *(1 mole of Magnesium/2 moles of HCl)*(24.3g og Mg/mol)= g go Mg
I apologize
Isn't it two moles of HCl for one mole of Mg?
I screwed up. I completely read the problem wrong. The equation I wrote is correct which, I suppose, led to Devron correcting his response. I gave you the answer for mols HCl for 2 mol Mg and that wasn't the question. Sorry about that.
I was correcting it, and didn't know that you were correcting it. i went back and looked at what I did, and after looking at the question again, I noticed that I assumed it was a 1:1 mole reaction, isn't the case. When I refreshed the window, because sometimes you and the poster post a response while I type slowly, I noticed what you did and asked the question.