Answer:
25.6g de HF son producidos
Explanation:
<em>...¿Cuánto HF es producido?</em>
Para resolver este problema debemos convertir la masa de cada reactivo a moles usando su masa molar. Como la reacción es 1:1, el reactivo con menor número de moles es el reactivo limitante. Con las moles del reactivo limitante podemos obtener las moles de HF y su masa así:
<em>Moles CaF2:</em>
Masa molar:
1Ca = 40g/mol
2F = 19*2 = 38g/mol
40+38 = 78g/mol
50g CaF2 * (1mol/78g) = 0.641 moles CaF2
<em>Moles H2SO4:</em>
Masa molar:
2H = 2g/mol
1S = 32g/mol
4O = 64g/mol
98g/mol
100g H2SO4 * (1mol / 98g) = 1.02 moles H2SO4
Como las moles de CaF2 < Moles H2SO4: CaF2 es reactivo limitante.
<em>Moles HF usando la reacción:</em>
0.641 moles CaF2 * (2mol HF / 1mol CaF2) = 1.282 moles HF
<em>Masa HF:</em>
Masa molar:
1g/mol + 19g/mol = 20g/mol
1.282 moles HF * (20g/mol) =
<h3>25.6g de HF son producidos</h3>
Answer:
the hydrogen atom of one water molecule and the lone pair of electrons on an oxygen atom of a neighboring water molecule.
Answer:
a) 213.3 mg/L
b) 62.61 mg/L
c) 0.0225 mg/L
Explanation:
Theoretical oxygen demand (ThOD)is essentially the amount of oxygen required for the complete degradation of a given compound into the final oxidized products
a) Given:
Concentration of acetic acid,
= 200 mg/L


Based on the reaction stoichiometry:
mass of
= 60 g
mass of
= 2(32) = 64 g

b) Given:
Concentration of ethanol,
= 30 mg/L


Based on the reaction stoichiometry:
mass of
= 46 g
mass of
= 3(32) = 96 g

c) Given:
Concentration of sucrose,
= 50 mg/L


Based on the reaction stoichiometry:
mass of
= 342 g
mass of
= 12(32) = 384 g

169.32 grams of NH3 ........
Answer:
The estimated feed rate of logs is 14.3 logs/min.
Explanation:
The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.
That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.
Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.
To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

The specific gravity of the wood chips is 0.494.
The average volume of a log is

The weight of one log is

To provide 3617 ton/day of wood chips, we need


The feed rate of logs is 14.3 logs/min.