The balanced chemical reaction is:
<span>3N2H4(l)→4NH3(g)+N2(g)
</span>
The amounts given for the N2H4 reactant will be the starting point for our calculations.
2.6mol N2H4 ( 4 mol NH3 / 3 mol N2H4 ) = 3.47 mol NH3
4.05mol N2H4 ( 4 mol NH3 / 3 mol N2H4 ) = 5.4 mol NH3
63.8g N2H4 <span>( 4 mol NH3 / 3 mol N2H4 ) = 85.07 mol NH3</span>
Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.
2 S + 3 O₂ → 2 SO₃
The stoichiometric calculations is as follows:
7 g S * 1 mol/32.06 g S = 0.218 mol S
Moles O₂ needed = 0.218 mol S * 3 mol O₂/2 mol S = 0.3275 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.3275 mol O₂ * 32 g/mol = 10.48 g O₂
All of them are soluble salt.
First one dissociates into two ions.
The second one dissociates into 3 ions.
The third dissociate into 4 ions. therefore, Al(NO3)3
Answer:
Al 72.61%
Mg 27.39%
Explanation:
To obtain the mass percentages, we need to place the individual masses over the total mass and multiply by 100%.
If we observe clearly, we can see that the parameters given are the moles. We need to convert the moles to mass.
To do this ,we need to multiply the moles by the atomic masses. The atomic mass of aluminum is 27 while that of magnesium is 24.
Now, the mass of aluminum is thus = 27 * 0.0898 = 2.4246g
The mass of magnesium is 0.0381 * 24 = 0.9144g
We can now calculate the mass percentage.
The total mass is 0.9144 + 2.4246 = 3.339g
% mass of Al = 2.4246/3.339 * 100 = 72.61%
% mass of Mg = 0.9144/3.39 * 100 = 27.39%
I’m sorry but where’s the worksheet lol......
