Hydrogen and oxygen gas are mixed in a 7.75 flask at 65 degree C and contains 0.
ID: 479278 • Letter: H
Question
Hydrogen and oxygen gas are mixed in a 7.75 flask at 65 degree C and contains 0.472 grams of hydrogen and 4.68 grams of oxygen. Calculate the partial pressures of oxygen and hydrogen in the flask Calculate the total pressure as well Magnesium metal (0.100 mol)and a volume of aqueous hydrochloric acid that contains 0.400 mol of HCl are combined and react to completion as shown below. Calculate the volume of hydrogen gas produced in liters at STP conditions. Mg(s) + 2HCl (aq) rightarrow MgCl (aq) +H_2(g) (R = 0.08206 L middot atm/K middot mol)Explanation / Answer
Ans 1. Number of moles of H2 = mass of substance/molar mass
= 0.472/2 = 0.236
Number of moles of O2 = mass of substance/ molar mass
=4.68/32 = 0.146
Temperature = 650 C = 338K Volume= 7.75L
We know from ideal gas equation P=nRT/V
For H2 partial pressure p1 = nRT/V = 0.236 x 0.0821 x 338/7.75
= 0.84 atm
For O2 partial pressure p2 = nRT/V = 0.146 x 0.0821 x 338 /7.75
=0.522 atm
Total pressure= p1 + p2
= 0.84+0.522 =1.362 atm
Ans 2. For ideal gas condition
P= 1atm , n=1 ,R=0.0821 , T=273
V= nRT/P
=1 x 0.0821 x 273 / 1
=22.411 L
Now, The balanced equation is
Mg + 2HCl -> MgCl2 + H2
Given moles of Mg = 0.1 mol
Since the equation is balanced
therefore 1 mol of Mg gives 1 mol of H2
0.1 mol of Mg = 0.1 mol of H2
We proved above for STP
1 mol of ideal gas has volume=22.411L
0.1 mol of H2 = 22.411 x 0.1 =2.24 L of H2
Therefore Volume of H2 at STP=2.24L
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