Question #72f34

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Question #72f34

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Answer 1 · 2017-12-10T21:49:06

$[H C l] ≅ 0.065 \cdot m o l \cdot L^{- 1}$ $[HCl]~=0.065*mol*L^-1$

Explanation:

We first write the stoichiometric equation....

$H C l (a q) + K O H (a q) \to K C l (a q) + H_{2} O (l)$ $HCl(aq) + KOH(aq) rarr KCl(aq) +H_2O(l)$

...to establish the 1:1 equivalence...

And thus $[H C l] = \frac{25.81 \cdot m L \times 10^{- 3} \cdot L \cdot m L^{- 1} \times 0.1250 \cdot m o l \cdot L^{- 1}}{50.00 \times 10^{- 3} \cdot L}$ $[HCl]=(25.81*mLxx10^-3*L*mL^-1xx0.1250*mol*L^-1)/(50.00xx10^-3*L)$

$= 0.06453 \cdot m o l \cdot L^{- 1}$ $=0.06453*mol*L^-1$

Answer 2 · 2017-12-10T21:58:35

$0.065 M$ $"0.065 M"$

Explanation:

The formula for the reaction is

$HCl + KOH \to KCl + H_{2} O$ $"HCl" + "KOH" -> "KCl" + "H"_2"O"$

and its already balanced (yea!).

When the mL and M of something $(KOH)$ $("KOH")$ is givenm use cross multiplication to find how many mols (remember $M$ $"M"$ is $mol/liter$ $"mol/liter"$ , or $mol/1000 mL$ $"mol/1000 mL"$ ):

$\frac{0.1250 mol}{1000 mL} = \frac{x mol}{25.81 mL}$ $"0.1250 mol"/"1000 mL" = "x mol"/ "25.81 mL"$

to get $3.23 \times 10^{- 3}$ $3.23xx10^-3$ $moles KOH$ $"moles KOH"$ , and since in the reaction equation all coefficients are 1, that means there were $3.23 \times 10^{- 3}$ $3.23xx10^-3$ $moles$ $"moles"$ of $HCl$ $"HCl"$ in the $50 mL$ $"50 mL"$ sample.

Then, you use cross multiplication again to find how many moles would be in $1000 mL$ $"1000 mL"$ of that sample:

$\frac{x mol}{1000 mL} = \frac{3.23 \times 10^{- 3} mol}{50 mL}$ $"x mol"/"1000 mL" = (3.23xx10^-3 "mol")/ "50 mL"$

which will give you $0.065 mol/1000 mL$ $"0.065 mol/1000 mL"$ , which is $0.065 M$ $"0.065 M"$ .

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Question #72f34

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