Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Nabz on November 03, 2007, 05:50:00 AM
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Hi, all
theres an AS Level question in my book that i am really stuck with.
it says:
Soda lime is 85% NaOH and 15% CaO. What volume of 0.500M HCl (aq) is needed to neutralize 2.50 g of soda lime? [HINT: consider each reaction seperately]
hope thats clear enough
thanks
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Write reaction equations. How much CaO and how much NaOH per 2.5 g of mixture?
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Write reaction equations. How much CaO and how much NaOH per 2.5 g of mixture?
i dnt get what you mean by "Write reaction equations" but this is what i think:
because i am given the % abundances, maybe you have to find the empirical formula?
so
NaOH + CaO = soda lime
85/(23+16+1) 15/ (40.1+ 16)
2.125/0.267 0.267/0.267
ratio= (NaOH) 8 : (Cao) 1
am i on the right path?
or do you just simply find the % from the mass given- 15% of 2.5g = 0.375g (CaO)
- 85% of 2.5g= 2.125g (NaOH)
:-\ confused
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You have two separate reactions taking place - two separate neutralizations. As in almost every stoichiometry problem you have to write balanced reaction equations to be able to calculate amounts of substances.
At this point you know there is 0.375 g CaO - how much HCl needed for reaction? 2.125 g NaOH - how much HCl needed for neutralization?
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You have two separate reactions taking place - two separate neutralizations. As in almost every stoichiometry problem you have to write balanced reaction equations to be able to calculate amounts of substances.
At this point you know there is 0.375 g CaO - how much HCl needed for reaction? 2.125 g NaOH - how much HCl needed for neutralization?
OK, so two separate reactions, so 2 seperate answers?
this is what i got:
n=m/Mr, so 0.375g/56.1 = 0.00668. ratio= CaO 1: HCl 1, so n(HCl)= 0.00668
m(of HCl)= n x Mr: 0.00668x36.5 = 0.244
n(of NaOH)=m/Mr, so 2.125g/40 = 0.0531. ratio= NaOH 1: HCl 1, so n(HCl)= 0.0531
m(of HCl)= n x Mr: 0.0531x36.5 = 1.938g
so i have found the mass in this attempt
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here i have found the Volume
V=n/conc.
n(CaO)=m/Mr, so 0.375g/56.1 = 0.00668. ratio= CaO 1: HCl 1, so n(HCl)= 0.00668
V=n/conc. so 0.00668x0.500M = 0.00334x1000= 3.34cm^3 Volume needed
n(of NaOH)=m/Mr, so 2.125g/40 = 0.0531. ratio= NaOH 1: HCl 1, so n(HCl)= 0.0531
V=n/conc. so 0.0531x0.500M = 0.002655x1000= 26.55cm^3 Volume needed
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did i have to find the mass needed for HCl or the Volume needed for HCl when you asked me "how much HCl needed for reaction?"
thanks
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CaO 1: HCl 1
You don't know this yet. You have to:
write balanced reaction equations to be able to calculate amounts
OK, so two separate reactions, so 2 seperate answers?
You add the 2 answers together to get 1 answer.
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2.50g soda lime * .85 = 2.215 g NaOH * 1 mole / 40g = .055 mole NaOH
.500M HCl = .5 mole/ Liter
.055 moles NaOh/.5 moles HCL * .055 moles HCL/x liters = .11 liters HCL, or 110 ml HCL(aq)