Answer is: enthalpy for this reaction is -175.91 kJ.
Chemical reaction: NH₃ + HCl → NH₄Cl.
ΔrH = ∑ΔfH(products of reaction) - ∑ΔfH(reactants).
ΔrH = ΔfH(NH₄Cl) - (ΔfH(NH₃) + ΔfH(HCl)).
ΔrH = -314.4 kJ/mol · 1 mol - (-46.19 kJ/mol · 1 mol + (-92.30 kJ/mol) · 1 mol).
ΔrH = -314.4 kJ + 138.49 kJ.
ΔrH = -175.91 kJ.
Answer:
3
Explanation:
3 ampers (it says in the problem)
The ore contains 55.4% calcium phosphate (related to the mineral apatite) so the amount of Ca3(PO4)2 is 55.4%x=1000g so x=1000/0.554= 1.805kg. Now for the % of P in this amount of calcium phosphate, use all the masses of the elements in Ca3PO4= Ca=40.078 x 3= 120.23 and (PO4)2= (30.974+64)2=189.95 (NB oxygen is 16 mass x 4 =64) so the total mass is 310.2 and we have 61.95 of P (Pmass x 2) so 61.95/3102.= 0.19 or 19% P. So of the 1.805 x 0.19= 0.34kg of phosphorus.
