Chemistry Problem help Please?
A 5.55g sample of an aqueous solution of HCl contains an unknown amount of the acid. If 37.8 mL of 0.534 M NaOH is required to neutralize the HCl, what is the percent by weight of the HCl acid in the mixture?
Please show work, I really need to know how to do this type of problem before my test, ty for your Help.
Answer:
37.8 ml of 0.534 M NaOH has
37.8ml * 0.534 moles/1000ml NaOH = 0.0202 mole NaOH
NaOH(aq) + HCl(aq) --> NaCl(aq) + H2O(l)
1:1 relationship
so you need 0.0202 moles HCl in the 5.55 g sample
MW = 36.5 g/mole
0.0202 moles * 36.5 g/mole HCl = 0.737 g
so the weight % would then be:
0.737 g/ 5.55g * 100 = 13.2 %
HCl + NaOH --> NaCl + H2O
moles HCl = moles NaOH
moles HCl = M(NaOH) X V(NaOH)
moles HCl = 0.534 M X 0.0378 L = 0.0202 moles
0.0202 moles X 36.45 g HCl/mol HCl = 0.736 g HCl
% HCl = (0.736 g/5.55 g) X 100 = 13.3 %
The neutralisation:
HCl + NaOH = NaCl + H2O
from the equation we read that
n(HCl) = n(NaOH)
Let's calculate:
n(HCl) = c(NaOH)×V(NaOH)
m(HCl) = n(HCl)×M(HCl)
since n(HCl)=n(NaOH), you can replace N(HCl) with n(NaOH)
m(HCl)=n(NaOH)×M(HCl) = c(NaOH)×V(NaOH)×M(HCl)
W = m(HCl) / m(sm)
W = V(NaOH)×c(NaOH)×M(HCl) / m(sm)
W = 0,0378L × 0,534 mol/L × 35,5 g/mol / 5,5g
W = 0,13 = 13%
n=moles
c=molarity
V=volume
M=mole mass, which is 35,5 g/mol for HCl
W = mass percentage
m(sm) =mass of the sample
Since it is a complete neutralisation reaction,
no. of equivalents of HCl = no. of eq. of NaOH
given wt. / eq. wt. (of HCl) = Normality * Volume in litres (of NaOH)
Normality = Molarity in case of NaOH
so, equation becomes
x / 36.5 = 0.534 * 37.8 / 1000
x = 0.7367598 g
% by weight
= x / 5.55 * 100 %
= 13.27 %
Hope it is rite... 4 any help, u can check out the foll. website:
I try to do my best.
1) First, we must remember that this a titration made on acid by a standard basic solution (NaOH). Hence, the formula we must use is:
M1V1 = M2V2
where "1" refers to acid and "2" refers to base.
M1V1 = (0.534 M)(0.0378 L)
M1V1 = 0.0201 moles of acid
We know that molecular weight of HCl is 36.5 g/mol, then:
mass HCl = (0.0201 moles)(36.5 g/mol) = 0.7367 grams
Percent (w/w) of HCl in sample = (0.7367 g / 5.55 g)x100%
= 13.27%
That's all !!
Good luck!
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Please show work, I really need to know how to do this type of problem before my test, ty for your Help.
Answer:
37.8 ml of 0.534 M NaOH has
37.8ml * 0.534 moles/1000ml NaOH = 0.0202 mole NaOH
NaOH(aq) + HCl(aq) --> NaCl(aq) + H2O(l)
1:1 relationship
so you need 0.0202 moles HCl in the 5.55 g sample
MW = 36.5 g/mole
0.0202 moles * 36.5 g/mole HCl = 0.737 g
so the weight % would then be:
0.737 g/ 5.55g * 100 = 13.2 %
HCl + NaOH --> NaCl + H2O
moles HCl = moles NaOH
moles HCl = M(NaOH) X V(NaOH)
moles HCl = 0.534 M X 0.0378 L = 0.0202 moles
0.0202 moles X 36.45 g HCl/mol HCl = 0.736 g HCl
% HCl = (0.736 g/5.55 g) X 100 = 13.3 %
The neutralisation:
HCl + NaOH = NaCl + H2O
from the equation we read that
n(HCl) = n(NaOH)
Let's calculate:
n(HCl) = c(NaOH)×V(NaOH)
m(HCl) = n(HCl)×M(HCl)
since n(HCl)=n(NaOH), you can replace N(HCl) with n(NaOH)
m(HCl)=n(NaOH)×M(HCl) = c(NaOH)×V(NaOH)×M(HCl)
W = m(HCl) / m(sm)
W = V(NaOH)×c(NaOH)×M(HCl) / m(sm)
W = 0,0378L × 0,534 mol/L × 35,5 g/mol / 5,5g
W = 0,13 = 13%
n=moles
c=molarity
V=volume
M=mole mass, which is 35,5 g/mol for HCl
W = mass percentage
m(sm) =mass of the sample
Since it is a complete neutralisation reaction,
no. of equivalents of HCl = no. of eq. of NaOH
given wt. / eq. wt. (of HCl) = Normality * Volume in litres (of NaOH)
Normality = Molarity in case of NaOH
so, equation becomes
x / 36.5 = 0.534 * 37.8 / 1000
x = 0.7367598 g
% by weight
= x / 5.55 * 100 %
= 13.27 %
Hope it is rite... 4 any help, u can check out the foll. website:
I try to do my best.
1) First, we must remember that this a titration made on acid by a standard basic solution (NaOH). Hence, the formula we must use is:
M1V1 = M2V2
where "1" refers to acid and "2" refers to base.
M1V1 = (0.534 M)(0.0378 L)
M1V1 = 0.0201 moles of acid
We know that molecular weight of HCl is 36.5 g/mol, then:
mass HCl = (0.0201 moles)(36.5 g/mol) = 0.7367 grams
Percent (w/w) of HCl in sample = (0.7367 g / 5.55 g)x100%
= 13.27%
That's all !!
Good luck!
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