It's all three of the answers
Answer:
son 12.6 gramos de HF
Explanation:
Tienes que saber qual es el reactor limitante en este caso es fluoruro con los 20 gramos puedes producer .631 mol qual son 12.6 gramos
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.
Answer:
Amount of Ca(NO3)2 produced = 14.02 g
Explanation:
The given reaction can be depicted as follows:
Ca(OH)2 + 2HNO3 → Ca(NO3)2 + 2H2O
Since it is given that HNO3 is in excess, the limiting reactant is Ca(OH)2
Now, Mass of Ca(OH)2 = 6.33 g
Molar mass of Ca(OH)2 = 74 g/mol

Based on the reaction stoichiometry:
1 mole of Ca(OH)2 forms 1 mole of Ca(NO3)2
Therefore, moles of Ca(NO3)2 produced from the moles of Ca(OH)2 reacted = 0.0855 moles
Molar mass of Ca(NO3)2 = 164 g/mol
