Why does it not matter what mass of NaOH is used to solve for q?
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So we recently did a lab in Chem to prove Hess's law. The reactions we performed were
NaOH+HClNaOH(s)+HClNaOH(s)⟶NaCl+H2O⟶NaCl+H2O⟶Na++OH−(1)(2)(3)(1)NaOH+HCl⟶NaCl+HX2O(2)NaOH(s)+HCl⟶NaCl+HX2O(3)NaOH(s)⟶NaX++OHX−
One of our post lab questions asks us to describe what would happen to the qq and ΔHΔH of reaction (3)(3) if twice the mass of NaOH(s)NaOH(s) were used.
For the third reaction, 4.1 degrees was the temperature change, 2 grams of NaOH(s)NaOH(s) were used, and 50 mL50 mL of water were used. This gave me
q=(52 g×4.18×4.1)=−891.176 Jq=(52 g×4.18×4.1)=−891.176 J
and
ΔH=(−891.176 J/0.05 mol)=−17824 J/molΔH=(−891.176 J/0.05 mol)=−17824 J/mol
When I used twice the mass of NaOH(s)NaOH(s) like the problem asked, my results were
q=(54 g×4.18×4.1)=−925.452 Jq=(54 g×4.18×4.1)=−925.452 J
and
ΔH=(−925.452/0.1 mol)=−9225.908 J/molΔH=(−925.452/0.1 mol)=−9225.908 J/mol
Considering my results, why does the very next problem say that in reaction (3)(3), it doesn't really matter what mass of NaOHNaOH is used? Why would it not matter? My ΔHΔH and qq values were different. I'm convinced that I made an error in my calculations, so I'd greatly appreciate some clarification here.
Also, it asks why is it important that 2 g2 g of NaOHNaOHbe used in reaction (2)(2)? Is it because it's reacting with HClHCl instead of water like in reaction (3)(3)?
NaOH+HClNaOH(s)+HClNaOH(s)⟶NaCl+H2O⟶NaCl+H2O⟶Na++OH−(1)(2)(3)(1)NaOH+HCl⟶NaCl+HX2O(2)NaOH(s)+HCl⟶NaCl+HX2O(3)NaOH(s)⟶NaX++OHX−
One of our post lab questions asks us to describe what would happen to the qq and ΔHΔH of reaction (3)(3) if twice the mass of NaOH(s)NaOH(s) were used.
For the third reaction, 4.1 degrees was the temperature change, 2 grams of NaOH(s)NaOH(s) were used, and 50 mL50 mL of water were used. This gave me
q=(52 g×4.18×4.1)=−891.176 Jq=(52 g×4.18×4.1)=−891.176 J
and
ΔH=(−891.176 J/0.05 mol)=−17824 J/molΔH=(−891.176 J/0.05 mol)=−17824 J/mol
When I used twice the mass of NaOH(s)NaOH(s) like the problem asked, my results were
q=(54 g×4.18×4.1)=−925.452 Jq=(54 g×4.18×4.1)=−925.452 J
and
ΔH=(−925.452/0.1 mol)=−9225.908 J/molΔH=(−925.452/0.1 mol)=−9225.908 J/mol
Considering my results, why does the very next problem say that in reaction (3)(3), it doesn't really matter what mass of NaOHNaOH is used? Why would it not matter? My ΔHΔH and qq values were different. I'm convinced that I made an error in my calculations, so I'd greatly appreciate some clarification here.
Also, it asks why is it important that 2 g2 g of NaOHNaOHbe used in reaction (2)(2)? Is it because it's reacting with HClHCl instead of water like in reaction (3)(3)?
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