Reacting mass calculation
Hi everyone,
I was doing a reacting mass question and realised that I had the answer wrong. But I didn't understand why. The question is attached, as well as my working, so if someone could tell me where I went wrong, I'd be thoroughly happy.
Posted from TSR Mobile
I was doing a reacting mass question and realised that I had the answer wrong. But I didn't understand why. The question is attached, as well as my working, so if someone could tell me where I went wrong, I'd be thoroughly happy.
Posted from TSR Mobile
Molar mass of nitric acid is NOT 126.
You used stoichiometric ratio twice - once assuming molar mass to be 126, once dividing number of moles by two. These are different approaches, you can't mix them.
Calculate number of moles of nitric acid using its true molar mass (63), calculate - from the stoichiometry - number of moles of nitrate produced, calculate mass of this nitrate.
You used stoichiometric ratio twice - once assuming molar mass to be 126, once dividing number of moles by two. These are different approaches, you can't mix them.
Calculate number of moles of nitric acid using its true molar mass (63), calculate - from the stoichiometry - number of moles of nitrate produced, calculate mass of this nitrate.
Original post by Borek
Molar mass of nitric acid is NOT 126.
You used stoichiometric ratio twice - once assuming molar mass to be 126, once dividing number of moles by two. These are different approaches, you can't mix them.
Calculate number of moles of nitric acid using its true molar mass (63), calculate - from the stoichiometry - number of moles of nitrate produced, calculate mass of this nitrate.
You used stoichiometric ratio twice - once assuming molar mass to be 126, once dividing number of moles by two. These are different approaches, you can't mix them.
Calculate number of moles of nitric acid using its true molar mass (63), calculate - from the stoichiometry - number of moles of nitrate produced, calculate mass of this nitrate.
Wow, I never knew that: I thought you had to apply to ratio to both. I will remember that now - thanks a lot!
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