Is the Calculation for Additional Moles of D2 Correct?

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SUMMARY

The calculation for the additional moles of D2 is confirmed to be correct, resulting in the need to add +10/(18.9) moles of D2, which equates to a mass of 2.1 grams. The discussion utilized the molar masses of H2 (2u), HD (3u), and D2 (4u) to derive the total moles equation. The total moles (nTot) were calculated as 10/2.7, leading to the conclusion that x, the additional moles of D2, is accurately represented as nTot/7.

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Homework Statement
All below
Relevant Equations
All below
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mx remets to mass in gram of the compost x
nx remets to moles of the compost x
nTot remets to the total moles
x remets to the additional moles of D2
I considered the molar mass:
MH2 = 2u
MHD = 3u
MD2 = 4u

10 = mH2 + mHD + mD2
10 = nH2*2 + nHD*3 + nD2*4
10 = nTot + 0.9nTot + 0.8nTot
nTot = 10/2.7

So:
0.3 = (nD2 + x)/nTot + x
x = nTot/7
x = 10/(18.9)

That is, we need to add + 10/(18.9) moles of D2,
10/(18.9) = m/4
m = 2.1 g

This seems right?
 
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LCSphysicist said:
Homework Statement:: All below
Relevant Equations:: All below

View attachment 267794

mx remets to mass in gram of the compost x
nx remets to moles of the compost x
nTot remets to the total moles
x remets to the additional moles of D2
I considered the molar mass:
MH2 = 2u
MHD = 3u
MD2 = 4u

10 = mH2 + mHD + mD2
10 = nH2*2 + nHD*3 + nD2*4
10 = nTot + 0.9nTot + 0.8nTot
nTot = 10/2.7

So:
0.3 = (nD2 + x)/nTot + x
x = nTot/7
x = 10/(18.9)

That is, we need to add + 10/(18.9) moles of D2,
10/(18.9) = m/4
m = 2.1 g

This seems right?
That's what I get. Nice job.
 
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