Asked by Justin
A Sample of M2O3 weighing 11.205g is converted to 15.380g of MCL3, where M is the unknown metal, then what is the identity of M?
Answers
Answered by
bobpursley
from M2O3, valence is +3
Now M2O3 converts to two moles of MCl3.
M2O3 + Cl2 >> 2MCl3 + ???
so 11.205 grams of M2O3 is x moles, and 15.380g of MCl3 is 2x moles.
mole mass MCl3= M + 3*35=M+105
mole mass M2O3= 2M+3*16=2M+48
moles of MCl3=15.380/(M+105)=2x
mass of M in M2O3=11.205/(2M+48)=x
from these, we can set 2x=2x or
15.380/(M+105)=2*11.205/(2M+48)
now solve for M.
15.380(2M+48)=22.41(M+105)
30.76M+2304=22.41M+2353
8.3M=49
M=5.9n, where n is an integer. Looks like Boron to me
Lets check if it is Boron
then reactant is moles=11.205/(10.8*2+3*48)
moles=.068moles
and moles of product is
moles=15.38/(10.8+3*35)=.138
which is indeed twice the moles as the reactant, as expected.
Now M2O3 converts to two moles of MCl3.
M2O3 + Cl2 >> 2MCl3 + ???
so 11.205 grams of M2O3 is x moles, and 15.380g of MCl3 is 2x moles.
mole mass MCl3= M + 3*35=M+105
mole mass M2O3= 2M+3*16=2M+48
moles of MCl3=15.380/(M+105)=2x
mass of M in M2O3=11.205/(2M+48)=x
from these, we can set 2x=2x or
15.380/(M+105)=2*11.205/(2M+48)
now solve for M.
15.380(2M+48)=22.41(M+105)
30.76M+2304=22.41M+2353
8.3M=49
M=5.9n, where n is an integer. Looks like Boron to me
Lets check if it is Boron
then reactant is moles=11.205/(10.8*2+3*48)
moles=.068moles
and moles of product is
moles=15.38/(10.8+3*35)=.138
which is indeed twice the moles as the reactant, as expected.
Answered by
DrBob222
Check this step.
hen reactant is moles=11.205/(10.8*2+3*48)
<b> Is that 3*16?</b>
hen reactant is moles=11.205/(10.8*2+3*48)
<b> Is that 3*16?</b>
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