Answer:
280.8 g
Explanation:
Definimos la reaccion:
2NaOH + FeSO₄ → Na₂SO₄ + Fe(OH)₂
Como tenemos la masa de NaOH, asumimos que el sulfato de hierro (II) es el reactivo en exceso.
Definimos masa de reactivo: 250 g . 1mol / 40g = 6.25 mol
2 moles de NaOH producen 1 mol de hidroxido ferroso
Entonces 6.25 moles producirán, la mitad (6.25 . 1) /2 = 3.125 moles
Convertimos los moles a masa:
3.125 mol . 89.85 g/mol = 280.8 g
You must remember that oxidation number of hydrogen in acids is always +1, oxidation number of oxygen in oxides & acids is always -2... metals has always oxidation number on plus!
group NO3 comes from HNO3...and oxidation number of whole acid group is always on minus and equal to the amount of hydrogen atoms in this acid... so oxidation number of NO3 = -1
we have 2 NO3 groups so 2*(-1) = -2 and that is the reason why oxidation number of Fe in this formula must be +2... because sum of all elements always gives 0!
Now we could count of oxidation number for nitrogen... we write HNO3 and start counting from right to left:
3*(-2) from oxygens + 1 from hydrogen = -5
so nitrogen must have +5 oxidation number... because sum all in formula must be 0.
984 grams of strontium will be recovered from 9.84x10^8 cubic meter of seawater.
Explanation:
From the question data given is :
volume of strontium in sea water= 9.84x10^8 cubic meter
(1 cubic metre = 1000000 ml)
so 9 .84x10^8 cubic meter
= 984 ml.
density of sea water = 1 gram/ml
from the formula mass of strontium can be calculated.
density = 
mass = density x volume
mass = 1 x 984
= 984 grams of strontium will be recovered.
98400 centigram of strontium will be recovered.
Strontium is an alkaline earth metal and is highly reactive.
