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3 years ago

Which of the following is a correct balanced equation for the neutralization reaction that occurs

Chemistry

1 answer:

PtichkaEL [24]3 years ago

3 0

Neutralization reaction takes place between an acid and a base to form salt and water.

In the given reaction the acid is hydrochloric acid HCl and the base is aluminum hydroxide (Al(OH) $_{3}$ ).

The correct balanced equation for the neutralization reaction that occurs between aluminum hydroxide and hydrochloric acid can be represented as:

$Al(OH)_{3}(aq)+ 3HCl(aq) --->AlCl_{3}(aq)+3H_{2}O(l)$

Al(OH) $_{3}$ can give 3 hydroxide, $OH^{-}$ ions which can neutralize the 3 $H^{+}$ ion produced by 3 HCl molecules giving 3 mole of water and the salt aluminum chloride, AlCl $_{3}$ .

Therefore the correct option will be.

D. Al(OH)3 + 3 HCl (aq) --> AlCl3(aq) + 3 H2O(l)

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Hydrogen gas can be produced by reacting aluminum with sulfuric acid. how many moles of sulfuric acid are needed to completely r

g100num [7]

The balanced equation for the reaction is as follows
2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂
stoichiometry of Al to H₂SO₄ is 2:3
number of Al moles reacted - 15.0 mol
if 2 mol of Al react with 3 mol of H₂SO₄
then 15.0 mol of Al reacts with - 3/2 x 15.0 mol = 22.5 mol
22.5 mol of H₂SO₄ is required

8 0

3 years ago

Read 2 more answers

Find the equilibrium value of [CO] if Kc=14.5 : CO (g) + 2H2 (g) ↔ CH3OH (g) Equilibrium concentrations: [H2] = 0.322 M and [CH3

Annette [7]

Answer:

The equilibrium value of [CO] is 1.04 M

Explanation:

Chemical equilibrium is the state to which a spontaneously evolving chemical system, in which a reversible chemical reaction takes place. When this situation is reached, it is observed that the concentrations of substances, both reagents and reaction products, they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.

Reagent concentrations and products in equilibrium are related by the equilibrium constant Kc. Being:

aA + bB ⇔ cC + dD

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }$

Then this constant Kces equals the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

In this case:

$Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }$

You know:

Kc= 14.5
[H₂]= 0.322 M
[CH₃OH] =1.56 M

Replacing:

$14.5=\frac{1.56}{[CO]*0.322^{2} }$

Solving:

$[CO]=\frac{1.56}{14.5*0.322^{2} }$

[CO]= 1.04 M

The equilibrium value of [CO] is 1.04 M

8 0

3 years ago

PI3Cl2 is a nonpolar molecule. Based on this information, determine the I−P−I bond angle, the Cl−P−Cl bond angle, and the I−P−Cl

alexdok [17]

Answer: Yes I-P-Cl = 90

Explanation:

This is because the angle formed between I-P-Cl is perpendicular hence the angle is 90°

3 0

3 years ago

A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.

BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

$\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\ &=185.59\text{ g/mol} \end{aligned}$

Dimensional Analysis:

$\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }$

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

8 0

3 years ago

Periodic trends are examined by looking down the _ of elements or by looking across the _ of the elements

nikklg [1K]

Answer: yo sorry this a hard one

Explanation:

bro

6 0

3 years ago