First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.
PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂
The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = <em>11.69 g</em>
1.5052g BaCl2.2H2O => 1.5052g / 274.25 g/mol = 0.0054884 mol
=> 0.0054884 mol Ba
<span>This means that at most 0.0054884 mol BaSO4 can form since Ba is the limiting reagent. </span>
<span>0.0054884 mol BaSO4 => 0.0054884 mol * 233.39 g/mol = 1.2809 g BaSO4</span>
The total energy can be found by adding the different energies:
628 + 15,600 + 712
= 16.94 kJ
Answer:
In 1000 ml there is 0.10 moles of Fe 2+
Therefore, in 10 ml there is (0.1/1000)*10= 0.001 mol of Fe2+
mole ratio for rxn Fe2+ : MnO4- is
1 : 2
therefore if 0.001 moles of Fe2+ react then 0.001*2 =0.002 moles of MnO4- react with Fe2+
hence, molarity of MnO4- = (mol*vol)/1000
= 0.002*10.75/1000= 2.15*10-5M
Explanation:
Hope this helps
Answer:
The mass of Mg consumed is 21.42g
Explanation:
The reaction is
As per balanced equation, three moles of Mg will react with one mole of nitrogen to give one mole of magnesium nitride.
as given that mass of nitrogen reacted = 8.33g
So moles of nitrogen reacted = 
moles of Mg required = 3 X moles of nitrogen taken = 3X0.2975 = 0.8925mol
Mass of Mg required = moles X molar mass = 0.8925 X 24 = 21.42 g
