How many moles of HCl should be added to Zn to get 20g H?
Reaction: Zn + 2HCl = ZnCl2 + H2[/code]
Please help me to solve this chemistry problem
Moderators: Xen, expert, ChenBeier
Ok lets begin the math:
First we se that 1 mole of Zn and 2 moles oh HCl gives us 1 mole oh hydrogen.
1 mole oh hydrogen weights around 2 grams seeing as molar mass oh 1 hydrogen atom is around 1.
There for we just devide the mass of reguired hydrogen by the molar mass of 1 mol and we get:
20g/(2g/mol)= 10 mol H2
And now since we needed 2 moles of HCl to make 1 mole of H2 we just multiply the moles from up there by 2 getting us:
10mol*2= 20 moles of HCl
There you go hopefully it helped!
First we se that 1 mole of Zn and 2 moles oh HCl gives us 1 mole oh hydrogen.
1 mole oh hydrogen weights around 2 grams seeing as molar mass oh 1 hydrogen atom is around 1.
There for we just devide the mass of reguired hydrogen by the molar mass of 1 mol and we get:
20g/(2g/mol)= 10 mol H2
And now since we needed 2 moles of HCl to make 1 mole of H2 we just multiply the moles from up there by 2 getting us:
10mol*2= 20 moles of HCl
There you go hopefully it helped!
I guess, one very important condition is missing here - the mass of Zn.
If you have got Zn in quantity lesser than 65.38 grams (1 mole of Zn), any amount of HCl can't help you to produce needed quantity of H.
If you have exactly 65.38 grams of Zn, then ANY amount of HCl, equal or exceeding 36.45 x 2=72.9 grams, should produce only 20 grams of H.
If you have more than 65.38 grams of Zn, then you have to be careful and use exactly 72.9 grams of HCl to produce 20 grams of H.
This is my understanding...
If you have got Zn in quantity lesser than 65.38 grams (1 mole of Zn), any amount of HCl can't help you to produce needed quantity of H.
If you have exactly 65.38 grams of Zn, then ANY amount of HCl, equal or exceeding 36.45 x 2=72.9 grams, should produce only 20 grams of H.
If you have more than 65.38 grams of Zn, then you have to be careful and use exactly 72.9 grams of HCl to produce 20 grams of H.
This is my understanding...