Answer to Question #136977 in General Chemistry for Miça

Question

The uranium minerals found in nature must be refined and enriched in 235U before the uranium can be used as a fuel in nuclear reactors. One procedure for enriching uranium begins with the reaction of natural UO2 with HF to form UF4, which is then converted into UF6 by reaction with fluorine

UO2 + 4HF—>UF4 + 2H2O

UF4+F2–> UF6

how many kg of HF are needed to completely react with 4.01kg of UO2

How much UF6 can be produced from 670.5g of UO2

Answer 1

Solution:

The molar mass of UO₂ 267.04 gram/mol = 267.04 kg/kmol

The molar mass of UF₆ 349.04 gram/mol = 349.04 kg/kmol

The molar mass of HF 20.0 gram/mol = 20.0 kg/kmol

To express a known mass of matter as an amount is use the molar mass, M,

n = m/M

1. According to the reaction equation UO₂ + 4HF—>UF₄ + 2H₂O

for 1 kmol of UO₂ are needs 4 kmol of HF.

And then, calculate mass of HF

m HF = m of UO₂/M of UO₂ × n of HF × M of HF =  

$(4.01 kg 〖UO〗_2 )/(267.04 kg/kmol 〖UO〗_2 )×$$(4 kmol HF)/(1 kmol)×20.0 kg/kmol HF$ = 1.20 kg HF

2. According to the reaction equations

from 1 mol of UO₂ can be produced 1 mol of UF₆.

And then, calculate mass of UF₆

m UF₆ = m of UO₂ / M of UO₂ × n of UF₆ × M of UF₆

$= (670.5g 〖UO〗_2)/(267.04 g/mol 〖UO〗_2 )×$

$(1 mol 〖UF〗_6 )/(1 mol)×349.04 g/mol 〖UF〗_6$ = 876.39 g UF₆

Answer:

1. are needed 1.20 kg of HF

2. can be produced 876.39 grams of UF₆