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3 years ago

13

write the balanced equation for the reaction that occurs when butane undergoes combustionWhat is the coefficient of oxygen in th

e balanced equation?

1 answer:

love history [14]3 years ago

5 0

Answer:

2C4H10 + 13O2 ----> 4CO2 + 10H2O

The coefficient of oxygen in the balanced equation is 13

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What is the amount of aluminum chloride produced from 40 moles of chlorine and excess aluminum?

Vera_Pavlovna [14]

Answer: Molar mass of Al = 26.98 g/mol

mass of Al = 12 g

mol of Al = (mass)/(molar mass)

= 12/26.98

= 0.4448 mol

According to balanced equation

mol of AlCl3 formed = moles of Al

= 0.4448 mol

Answer: 0.445 mol

hope this help boo❤️❤️❤️

Explanation:

3 0

3 years ago

Using 18.8L, calculate the volume (in L) of water vapour that should be produced by the reaction of 35.0 g

victus00 [196]

Answer:

A. 18.8L

B. 75.2L of H2O.

Explanation:

A. Determination of the volume of 35g of C3H8.

Date obtained from the question include the following:

Mass of C3H8 = 35g

Temperature (T) = 40°C

Pressure (P) = 110KPa

Volume (V) =..?

Next, we shall determine the number of mole (n) in 35g of C3H8. This is illustrated below:

Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol

Mass of C3H8 = 35g

Mole of C3H8 =..?

Mole = mass /molar mass

Mole of C3H8 = 35/44

Mole of C3H8 = 0.795 mole

Finally, we shall determine the volume of 35g of C3H8 as follow:

Temperature (T) = 40°C = 40°C + 273 = 313K

Pressure (P) = 110KPa

Number of mole (n) = 0.795 mole

Gas constant (R) = 8.314 KPa.L/Kmol

Volume (V) =..?

PV = nRT

110 x V = 0.795 x 8.314 x 313

Divide both side by 110

V = (0.795 x 8.314 x 313)/110

V = 18.8L

Therefore, the volume of 35g of C3H8 under the conditions given is 18.8L

B. Determination of the volume of water vapour produced by the reaction of 35g of propane, C3H8.

From the calculations made in (A) above, 35g of C3H8 is equivalent to 18.8L of C3H8.

Thus, we can obtain the volume of water vapour produced as follow:

C3H8 + 5O2 —> 3CO2 + 4H2O

From the balanced equation above,

1L of C3H8 reacted to produce 4L of H2O.

Therefore, 18.8L of C3H8 will react to produce = (18.8 x 4)/1 = 75.2L of H2O.

Therefore, 75.2L of H2O were produced from the reaction.

4 0

3 years ago

What 2 properties make capillary action possible

Natali5045456 [20]

Cohesion and adhesion

7 0

3 years ago

Using the reaction below determine the amount of Sulfur proceduced in grams (this is a limiting reaction question) 2 H2S + SO2 -

marusya05 [52]

Answer:

<u>= 2.2 g pf S. produced</u>

Explanation:

Balanced Reaction equation:

$2H_{2} S + SO_{2}$ → $3S + 2H_{2} O$

1 mole of H2S - 34.1g

? moles - 3.2g

= 3.2/34.1 =<u> 0.09 moles of H2S</u>

Also,

1 mole of S02 - 64.07 g

? moles - 4.42g

= 4.42/64.07 <u>= 0.069 moles of SO2</u>

<u />

<em>Meaning SO2 is the limiting reagent</em>

Finally, 3 moles of S - 32g of sulphur

0.069 mole = ? g of Sulphur

= 0.069 x 32

<u>= 2.2 g pf S.</u>

7 0

3 years ago

Under standard-state conditions, which of the following half-reactions occurs at the cathode during the electrolysis of aqueous

Slav-nsk [51]

Answer:

The following reaction will occur at cathode:

$Ni^{+2}(aq)+2e--->Ni(s)$

Explanation:

The two half reaction during electrolysis of aqueous nickel sulfate will be

a) anode reaction :

Water will undergo oxidation and will evolve oxygen gas at anode as shown in the given reaction:

$2H_{2}O(l) ----> O_{2}(g) + 4H^{+}(aq) + 4e$

b) Cathode reaction: The reduction of Nickel ion will occur by gain of two electrons as shown in the given equation:

$Ni^{+2}(aq)+2e--->Ni(s)$

Thus the overall reaction will be:

$2H_{2}O(l) + 2Ni^{+2}----> O_{2}(g) + 4H^{+}(aq) + 2Ni(s)$

6 0

3 years ago