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3 years ago

5

In the lab vinegar acetic acid was mixed with baking soda sodium bicarbonate to form dioxide gas and sodium acetate. What are th

e products?

1 answer:

SpyIntel [72]3 years ago

4 0

2CH3COOH +Na2CO3 ----> 2CH3COONa + H20 + CO2

know you can find what all products formed from his reaction

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What volume of Co2 (carbon (iv) oxide)

hram777 [196]

Answer:

2.1056L or 2105.6mL

Explanation:

We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:

Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol

Mass of Na2CO3 = 10g

Mole of Na2CO3 =.?

Mole = mass /molar mass

Mole of Na2CO3 = 10/106

Mole of Na2CO3 = 0.094 mole

Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:

Na2CO3 + 2HCl —> 2NaCl + H2O + CO2

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CO2.

Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.

Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:

1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.

Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L

Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL

7 0

4 years ago

Iron and oxygen form rust. How many moles of rust should be produced if 1

kakasveta [241]

Answer:

1.387 moles

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

4Fe + 3O2 —> 2Fe2O3

Step 2:

Determination of the number of mole of Fe in 155.321g of Fe. This can be achieved by doing the following:

Mass of Fe = 155.321g

Molar Mass of Fe = 56g/mol

Number of mole of Fe =?

Number of mole = Mass/Molar Mass

Number of mole of Fe = 155.321/56

Number of mole of Fe = 2.774 mol

Step 3:

Determination of the number of mole of rust (Fe2O3) produced. This is illustrated below:

From the balanced equation above,

4 moles of Fe produced 2 moles of Fe2O3.

Therefore, 2.774 moles of Fe will produce = (2.774 x 2)/4 = 1.387 moles of Fe2O3.

Therefore, 1.387 moles of rust (Fe2O3) is produced from the reaction

6 0

3 years ago

Read 2 more answers

A piece of wood and a piece of steel are at the same temperature; however, the steel feels hotter. This is because the steel has

Anika [276]

B- Greater thermal conductivity

6 0

3 years ago

Read 2 more answers

An area in the desert where water is found is known as an.

lbvjy [14]

Answer:

Oasis

Explanation:

An area in the desert that has water in it is called an oasis.

6 0

2 years ago

Aluminum sulfate, known as cake alum, has a wide range of uses, from dyeing leather and cloth to purifying sewage. In aqueous so

NISA [10]

Answer:

a) The chemical reaction is given as:

$6NaOH(aq)+Al_2(SO_4)_3\rightarrow 2Al(OH)_3(s)+3Na_2SO_4(aq)$

2.571 grams of aluminum hydroxide is precipitated.

Explanation:

a) The chemical reaction is given as:

$6NaOH(aq)+Al_2(SO_4)_3\rightarrow 2Al(OH)_3(s)+3Na_2SO_4(aq)$

b)

Moles of NaOH = n

Volume of the NaOH = 185.5 mL = 0.1855 L( 1 mL =0.001 L)

Molarity of the solution = 0.533 M

$n=0.533 M\times 0.1855 mL=0.09887 mol$

Moles of aluminum = n

15.8 g of aluminum sulfate per liter.

Volume of solution = 627 mL = 0.627 L (1 mL= 0.001 L)

Mass of aluminium sulfate in solution = 15.8 g/L × 0.627 L =9.9066 g

Moles of aluminum sulfate = $\frac{9.9066 g}{342 g/mol}=0.02897 mol$

Moles of NaOH = 0.09887 mol

According to reaction, 6 moles of NaOH reacts with 1 mole of aluminum sulfate, then 0.09887 moles of NaOH will recat with :

$\frac{1}{6}\times 0.09887 mol=0.01648 mol$

This means that sodium hydroxide moles are in limiting amount.So, amount of aluminum hydroxide will depend upon moles of sodium hydroxide.

According to reaction, 6 moles of sodium hydroxide gives 2 moles of aluminium hydroxide, then 0.09887 moles of sodium hydroxide will give :

$\frac{2}{6}\times 0.09887 mol=0.03296 mol$ of aluminum hydroxide

Mass of 0.03296 moles of aluminum hydroxide:

0.03296 mol × 78 g/mol = 2.571 g

2.571 grams of aluminum hydroxide is precipitated.

8 0

3 years ago