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Calculating moles from mass and molarity

Дата публикации: 11-04-2024 10:36:46



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TheePhysicsStudent
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Homework Statement
I was practising questions from a book printed in the 1980s, and I'm unsure where I went wrong with this Q, the answer is 68.2%
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I dont use equations for chemistry mole calculations, i just think my way through it so I don't know
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In your initial hand-written equation, you have “##NaHCO_3 \cdot Na_2CO_3##. This means you have implicitly assumed that you are starting with equal (molar) amounts of ##NaHCO_3## and ##Na_2CO_3##. It goes wrong from there.

Suppose the 0.500g of reactant contains x grams of ##NaHCO_3## and y grams, of ##Na_2CO_3##.

Can you set up 2 simultaneous equations for x and y and solve them?

Edit: ask for hints if you can't!

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Maybe this could reassure you or maybe not:
1 decimeter is 10 centimeters.
(careful! these two lines have a mistake.)
0.100 mole/(dm^3) * (10/1)^3(dm/cm)^3

100 mole/(cm)^3

------------------------------------
THAT above is obviously wrong. This is the kind of trouble I have when I do things strictly through keyboard and computer screen. My work later using pen and paper was bettter.

Should have that been like so:
0.100 (moles/dm^3)*(1/10)(dm/cm)(1/10)(dm/cm)(1/10)(dm/cm)
0.100(1/1000)(moles/cm^3)
0.000100 moles per cubic centimeter

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Maybe this could reassure you or maybe not:
1 decimeter is 10 centimeters.

0.100 mole/(dm^3) * (10/1)^3(dm/cm)^3

100 mole/(cm)^3

0.1 mole/1000cc=##0.0001\ moles/cc##

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