Chemistry limiting reaction
Determine the ACTUAL yeild of if 1.85 g Ca are reacted with 200. mL of 6.00 M HCl.
Assume:
M=molarity
Ca=40.08 g/mol
H=1.01g/mol
Cl=35.45 g/mol
The answer is 5.12.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Equation for the required chemical reaction:
Ca + 2HCl -> CaCl2 + H2
For every 1 mol of Ca to react, 2 mol of HCl is consumed and 1 mol of CaCl2 is formed.
200 mL of a 6.00 M HCl contains 1.2 mol HCl.
And 1.85 g Ca contains 0.046 mol Ca.
Hence, the maximum amount of HCl that can be utilised is 0.092 mol for after that there will remain no more Ca to react.
And since for every 1 mol Ca reacted, 1 mol CaCl2 is formed, in this case 0.046 mol CaCl2 will be formed which will amount to around 5.12 g.