When The balanced equation is:
2Al + 3CuCl2 ⇒3 Cu + 2AlCl3
So, we want to find the limiting reactant:
1- no. of moles of 2Al = MV/n = (Wt * V )/ (M.Wt*n*V) = Wt / (M.Wt *n)
where M= molarity, V= volume per liter and n = number of moles in the balanced equation.
by substitute:
∴ no. of moles of 2Al = 0.2 / (26.98 * 2)= 0.003706 moles.
2- no.of moles of 3CuCl2= M*v / n = (0.5*(15/1000)) / 3= 0.0025 moles.
So, CuCl2 is determining the no.of moles of the products.
∴The no. of moles of 3Cu = 0.0025 moles.
∴The no.of moles of Cu= 3*0.0025= 0.0075 moles.
and ∵ amount of weight (g)= no.of moles * M.Wt = 0.0075 * M.wt of Cu
= 0.0075 * 63.546 =0.477 g
Answer: The correct answer is -297 kJ.
Explanation:
To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:
2SO3 —> O2 + 2SO2 (196 kJ)
Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.
Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:
SO3 —>1/2O2 + SO2 (98 kJ)
S + 3/2O2 —> SO3 (-395 kJ)
Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.
S + O2 —> SO2
Now, we must add the enthalpies together to get our final answer.
-395 kJ + 98 kJ = -297 kJ
Hope this helps!
Answer:
The empirical formula is N2 CH4 O
Explanation:
Answer:
molality of sodium ions is 1.473 m
Explanation:
Molarity is moles of solute per litre of solution
Molality is moles of solute per kg of solvent.
The volume of solution = 1 L
The mass of solution = volume X density = 1000mL X 1.43 = 1430 grams
The mass of solute = moles X molar mass of sodium phosphate = 0.65X164
mass of solute = 106.6 grams
the mass of solvent = 1430 - 106.6 = 1323.4 grams = 1.3234 Kg
the molality = 
Thus molality of sodium phosphate is 0.491 m
Each sodium phosphate of molecule will give three sodium ions.
Thus molality of sodium ions = 3 X 0.491 = 1.473 m
