how many liters of hydrogen at 0 degrees Celsius and 1400 mmHg are produced if 15g of magnesium reacts with sulfuric acid?
and can you please explain how you got the answer because im not sure how to get it...
and can you please explain how you got the answer because im not sure how to get it...
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7.7 L of H2
So you are looking to find the liters of hydrogen: 1.) write and balance the equation, 2.) convert grams of magnesium to moles of hydrogen, and 3.) use the equation PV=nRT; where P= pressure (in atm), V= volume (in liters), n= moles of hydrogen, R= universal gas constant, and T= temperature (in kelvin)
1.) Mg + H2(SO4) ----> Mg(SO4) + H2
2.) 15.0g Mg x [1 mole / 24.3g] x [1 mole / 1mole] = .62 moles of H2
3.) PV=nRT
1,400 mmHg x [1 atm / 760 mmHg] = 1.8 atm
0*C + 273 = 273*K
Finally you can plug in the numbers and solve for the volume
(1.8 atm)(V) = (.62 moles)(.0821)(273*K)
V= (.62 moles)(.0821)(273*K) / (1.8 atm)
V= 7.7 L H2
So you are looking to find the liters of hydrogen: 1.) write and balance the equation, 2.) convert grams of magnesium to moles of hydrogen, and 3.) use the equation PV=nRT; where P= pressure (in atm), V= volume (in liters), n= moles of hydrogen, R= universal gas constant, and T= temperature (in kelvin)
1.) Mg + H2(SO4) ----> Mg(SO4) + H2
2.) 15.0g Mg x [1 mole / 24.3g] x [1 mole / 1mole] = .62 moles of H2
3.) PV=nRT
1,400 mmHg x [1 atm / 760 mmHg] = 1.8 atm
0*C + 273 = 273*K
Finally you can plug in the numbers and solve for the volume
(1.8 atm)(V) = (.62 moles)(.0821)(273*K)
V= (.62 moles)(.0821)(273*K) / (1.8 atm)
V= 7.7 L H2