Answer:
m(Na3AlF6)=57.6537 kg. Leftover mass will be 52.8644 kg
Explanation:
Al2O3(s)+6NaOH(l)+12HF(g)⟶2Na3AlF6+9H2O(g)
n(Al2O3)=m(Al2O3)/M(Al2O3)=14000/101.96=137.31 mol
n(NaOH)=m(NaOH)/M(NaOH)=59400/40=1485 mol
n(HF)=m(HF)/M(HF)=59400/20.01=2968 mol
To know what is in excess we divide moles by coefficients and compare numbers.
n(Al2O3) = 137.31 mol
n(NaOH)/6=247.5 mol
n(HF)/12=247.33 mol
so NaOH and HF are in excess
So
m(Na3AlF6)=n(Na3AlF6)*M(Na3AlF6)/1000=n(Al2O3)*2*M(Na3AlF6)/1000=57.6537 kg.
Leftover mass will be (n(NaOH)-n(Al2O3)*6)*40 + (n(HF)-n(Al2O3)*12)*20.01=26445.6+26418.8=52864.4 g= 52.8644 kg
Answer:
The pOH of a solution is the negative logarithm of the hydroxide-ion concentration. The pH of a solution can be related to the pOH.
I looked it up, so idk if it is right...
Answer: 313.6
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of require 3 moles of
Thus 2.8 moles of will require= of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 1 mole of give = 2 moles of
Thus 2.8 moles of give = of
Mass of
Theoretical yield of liquid iron is 313.6 g
Answer:
Explanation:
C. A proton and an electron
) adding phosphoric acid, said phosphoric acid reacting with said calcium carbonate to release gaseous carbon dioxide bubbles, and said water providing water of hydration to said plaster of Paris to form gypsum that sets around said carbon dioxide bubbles thereby forming said foamed plaster.