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Reacting mass calculations
What mass of hydrogen is produced when 1.00 g of sodium is reacted with excess water?
Could somebody just check my working out?
2Na and H2
Na - Divide mass by Mr to get moles = 1/23 = 0.043
There are 2 moles of Na so 0.043/2 = 0.0215 moles of hydrogen
Mass = Moles x Mr so o.o215 * 2 = 0.043
Is the mass of hydrogen 0.043 and is the working out right?
Could somebody just check my working out?
2Na and H2
Na - Divide mass by Mr to get moles = 1/23 = 0.043
There are 2 moles of Na so 0.043/2 = 0.0215 moles of hydrogen
Mass = Moles x Mr so o.o215 * 2 = 0.043
Is the mass of hydrogen 0.043 and is the working out right?
yes - 0.043 grams of hydrogen
So if the the question was mass of hydrogen when 100g of calcium is reacted with excess hydrochloric acid, would it be 5g?
2Ca + 2HCl ---> 2CaCl2 + H2
100g of Ca = 2.5 moles => 1.25 moles of H2 gas = 2.5g
EDIT: Ca + 2HCl ---> CaCl2 + H2 = 5g
100g of Ca = 2.5 moles => 1.25 moles of H2 gas = 2.5g
EDIT: Ca + 2HCl ---> CaCl2 + H2 = 5g
EierVonSatan
2Ca + 2HCl ---> 2CaCl2 + H2
100g of Ca = 2.5 moles => 1.25 moles of H2 gas = 2.5g
100g of Ca = 2.5 moles => 1.25 moles of H2 gas = 2.5g
Isn't the equation Ca + 2HCl ----> CaCl2 + H2?
Last question:
A mixture of anhydrous sodium carbonate and sodium hydrogencarbonate of mass 10 g was heated until it reached a constant mass of 8.708 g. Calculate the composition of the mixture in grams of each component. Sodium hydrogencarbonate thermally decomoses to form sodium carbonate.
Would the equation be Na2CO3 + NaHCO3 ---- > NaCO3 and then balance, find out moles of NaCO3, and then ....?
Thanks.
definite_maybe
Isn't the equation Ca + 2HCl ----> CaCl2 + H2?
lol sorry yes...I originally had water in there and forgot to alter it back
A mixture of anhydrous sodium carbonate and sodium hydrogencarbonate of mass 10 g was heated until it reached a constant mass of 8.708 g. Calculate the composition of the mixture in grams of each component. Sodium hydrogencarbonate thermally decomoses to form sodium carbonate.
Would the equation be Na2CO3 + NaHCO3 ---- > NaCO3 and then balance, find out moles of NaCO3, and then ....?
Thanks.
Would the equation be Na2CO3 + NaHCO3 ---- > NaCO3 and then balance, find out moles of NaCO3, and then ....?
Thanks.
You need to use 2NaHCO3 → Na2CO3 + H2O + CO2 - work out how much mass you've lost and relate it back to the starting amount of NaHCO3. Take away this mass from the original 10g to get how much Na2CO3 you started with.
EierVonSatan
You need to use 2NaHCO3 → Na2CO3 + H2O + CO2 - work out how much mass you've lost and relate it back to the starting amount of NaHCO3. Take away this mass from the original 10g to get how much Na2CO3 you started with.
I don't understand. How do I relate the mass lost to the starting amount?
definite_maybe
I don't understand. How do I relate the mass lost to the starting amount?
mass lost = 10 - 8.708 = 1.292 g = mass of H2O + CO2 given off
2NaHCO3 → Na2CO3 + H2O + CO2 in terms of Mr 2(84) → 106 + 18 + 44
we can treat H2O + CO2 as a single compound of mass 62
1.292/62 = 0.0208 mol - so this leads to 0.0416 mol of NaHCO3 at the start which is 3.5g
working backwards to check: 0.0416 mol of NaHCO3 → 0.0208 mol of Na2CO3 which is 2.205 g => 3.5 - 2.205 = 1.295 (close enough with rounding)
hope that makes sense...
EierVonSatan
hope that makes sense...
hope that makes sense...
Yep
It makes sense now. Thanks for all the help. 
EierVonSatan
mass lost = 10 - 8.708 = 1.292 g = mass of H2O + CO2 given off
2NaHCO3 → Na2CO3 + H2O + CO2 in terms of Mr 2(84) → 106 + 18 + 44
we can treat H2O + CO2 as a single compound of mass 62
1.292/62 = 0.0208 mol - so this leads to 0.0416 mol of NaHCO3 at the start which is 3.5g
working backwards to check: 0.0416 mol of NaHCO3 → 0.0208 mol of Na2CO3 which is 2.205 g => 3.5 - 2.205 = 1.295 (close enough with rounding)
hope that makes sense...
2NaHCO3 → Na2CO3 + H2O + CO2 in terms of Mr 2(84) → 106 + 18 + 44
we can treat H2O + CO2 as a single compound of mass 62
1.292/62 = 0.0208 mol - so this leads to 0.0416 mol of NaHCO3 at the start which is 3.5g
working backwards to check: 0.0416 mol of NaHCO3 → 0.0208 mol of Na2CO3 which is 2.205 g => 3.5 - 2.205 = 1.295 (close enough with rounding)
hope that makes sense...
Egad.
That helped a lot.
Danke schone.
Mask Of Sanity
Egad.
That helped a lot.
Danke schone.
That helped a lot.
Danke schone.
willkommen
This helped A LOT, thanks 7 years later ^^
No it doesn't make sense. Can you explain that again please? Thank you.
Original post by EierVonSatan
mass lost = 10 - 8.708 = 1.292 g = mass of H2O + CO2 given off
2NaHCO3 → Na2CO3 + H2O + CO2 in terms of Mr 2(84) → 106 + 18 + 44
we can treat H2O + CO2 as a single compound of mass 62
1.292/62 = 0.0208 mol - so this leads to 0.0416 mol of NaHCO3 at the start which is 3.5g
working backwards to check: 0.0416 mol of NaHCO3 → 0.0208 mol of Na2CO3 which is 2.205 g => 3.5 - 2.205 = 1.295 (close enough with rounding)
hope that makes sense...
2NaHCO3 → Na2CO3 + H2O + CO2 in terms of Mr 2(84) → 106 + 18 + 44
we can treat H2O + CO2 as a single compound of mass 62
1.292/62 = 0.0208 mol - so this leads to 0.0416 mol of NaHCO3 at the start which is 3.5g
working backwards to check: 0.0416 mol of NaHCO3 → 0.0208 mol of Na2CO3 which is 2.205 g => 3.5 - 2.205 = 1.295 (close enough with rounding)
hope that makes sense...
I still don't understand it. Can you explain it again please? Thank you.
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