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fomenos
2 years ago
6

You titrate 41.27 mL of 0.108 M Ca(OH)2 into 25.00 mL of citric acid (C6H307) (triprotic). What is the balanced equation and the

molarity of the acid?
Select both an equation and a molarity.
CoH2O2 (aq) + Ca(OH)2 (aq) + H20 (1) + Ca(C6H507)2 (aq)
2C6H807 (aq) + 3Ca(OH)2 (aq) + 6H20 (1) + Ca3(C6H507)2 (aq)
C6H2O7 (aq) + Ca(OH)2 (aq) + H20 (1) + CaC6H50% (aq)
3C6H307 (aq) + 2Ca(OH)2 (aq) + 6H20 (1) + Caz(C6H507)2 (aq)
0.267 M
0.178 M
0.0654 M
0.119 M
Chemistry
1 answer:
Triss [41]2 years ago
6 0

Answer:

3Ca(OH)2 + 2C6H8O7 → 6H2O + Ca3(C6H5O7)2

And 0.119M is the concentration of the citric acid.

Explanation:

In an acid-base reaction, the proton H+ and the hydroxil ion OH- reacts producing water. The ions of the acid and base (C6H5O7³⁻ and Ca²⁺ ions produce the respective salt) as follows:

Ca(OH)2 + C6H8O7 → H2O + Ca3(C6H5O7)2

To balance the Calcium ions:

3Ca(OH)2 + C6H8O7 → H2O + Ca3(C6H5O7)2

To balance the C6H5O7³⁻ ions:

3Ca(OH)2 + 2C6H8O7 → H2O + Ca3(C6H5O7)2

And to balance the oxygens of water:

3Ca(OH)2 + 2C6H8O7 → 6H2O + Ca3(C6H5O7)2

And this is the balanced reaction.

The moles of Ca(OH)2 that reacts are:

41.27mL = 0.04127L * (0.108mol/L) = 0.004457 moles Ca(OH)2

<em>Moles of citric acid:</em>

0.004457 moles Ca(OH)2 * (2mol C6H8O7 / 3mol Ca(OH)2) = 0.002971 moles C6H8O7

In 25.00mL = 0.02500L:

0.002971 moles C6H8O7 / 0.0250L =

<h3>0.119M</h3>

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2 years ago
A chemist must prepare of 800.0 ml potassium hydroxide solution with a pH of 13.00 at 25°.
ArbitrLikvidat [17]

Answer:

4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.

Explanation:

The pH of the solution = 13.00

pH + pOH = 14

pOH = 14 - pH = 14 - 13.00 = 1.00

pOH=-\log[OH^-]

1.00=-\log[OH^-]

[OH^-]=10^{-1.00} M=0.100 M

KOH(aq)\rightarrow K^+(aq)+OH^-(aq)

[KOH]=[OH^-]=[K^+]=0.100 M

Molariy of the KOH = 0.100 M

Volume of the KOH solution = 800 mL= 0.800 L

1 mL = 0.001 L

Moles of KOH = n

Molarity=\frac{Moles}{Volume(L)}

0.100 M=\frac{n}{0.800 L}

n = 0.0800 mol

Mass of 0.0800 moles of KOH :

0.0800 mol × 56 g/mol = 4.48 g

4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.

4 0
3 years ago
Imagine that you helped to discover a new element; A=302;Z=119. How many protons, electrons, and neutrons are in each atom it th
oee [108]

Explanation:

Given that,

Mass number, A = 302

Atomic number, Z = 119

We know that, atomic number = no of protons

Protons = 119

Mass no. = No. of neutrons + No. of protons

302 = No. of neutrons + 119

No. of neutrons = 302 - 119

= 183

No. of electrons = No. of protons

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5 0
2 years ago
139%
GaryK [48]

<u>Answer:</u>

The percent composition of this compound is 94%

<u>Explanation:</u>

The reaction can be formed as

2 \mathrm{Fe}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{FeCl}_{3}

\frac{\text { Weight of } \mathrm{Cl}_{2}}{\text { 3* Molar Mass of } \mathrm{Cl}_{2}}=\frac{\text { Weight of } \mathrm{Fe}}{2 * \text { Molar Mass of Fe }}

\frac{\text { Weight of } \mathrm{Cl}_{2}}{3 *(2 * 35.5)}=\frac{3.56}{2 * 55.8}

\text { Weight of } C l_{2}=\frac{3.56 * 3 * 71}{2 * 55.8}=6.79 \mathrm{g}

\mathrm{n}\left(\mathrm{Cl}_{2}\right)=\mathrm{m}\left(\mathrm{Cl}_{2}\right) / \mathrm{M}\left(\mathrm{Cl}_{2}\right)=6.79 / 71=0.1 \mathrm{m}

\mathrm{n}(\mathrm{Fe})=\mathrm{m}(\mathrm{Fe}) / \mathrm{M}(\mathrm{Fe})=3.56 / 55.8=0.06 \mathrm{m}

Based on no. of iron reacted,  

\mathrm{n}(\text { moles of } \mathrm{Fe})=\mathrm{n}\left(\text { moles of } \mathrm{FeCl}_{3}\right)

n = m/M

\mathrm{m}\left(\mathrm{FeCl}_{3}\right)=\mathrm{n}^{*} \mathrm{M}=0.06^{*} 162.5=9.75 \mathrm{g}

% composition ofFeCl_3  

=  (9.75 / 10.39)^{*} 100

= 94%

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3 years ago
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Lithium Bromine (LiBr) is formed from them....
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