Mg + 2HCl ---> MgCl2 + H2
What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl?
How much of the excess reagent in last problem is left over?
thanks in advance
2007-07-25 17:48:16 · 3 answers · asked by playpwnsu 2 in Science & Mathematics ➔ Chemistry
50.0g Mg is 2.058moles
75g HCl is 2.058moles; but you need 2 HCl for each Mg, so
1/2 or 1.027moles (24.96g of Mg left)
Also 1.027 moles of H2 produced or 23.0 liters [1.027 x 22.4l]
2007-07-25 18:26:47 · answer #1 · answered by Flying Dragon 7 · 0⤊ 0⤋
The number moles H2 produced is 1.0288 moles. You can probably figure that out this time.
Volume at STP is 1.0288*22.4L or 23.0L.
The excess is 25.0g of Mg. You can probably figure this out by this time.
2007-07-26 01:36:49 · answer #2 · answered by titanium007 4 · 0⤊ 0⤋
Molar mass of Mg = 24.30
Molar mass of HCL = 36.46
50.0g/24.30 = 2.057 mols x (1/1) = 2.057 mols product
75g /36.46 = 2.057 mols x (1/2)= 1.028 mols product
Therefore HCL is limiting reagent
you will be left with 1.028 mols of magnesium
2007-07-26 01:30:01 · answer #3 · answered by Anonymous · 0⤊ 0⤋