What are the classes of alkanol

50cm3 of a mixture of methane and hydrogen were mixed with excess oxygen and exploded, the product after cooling to the original

marin [14]

Answer:the initial composition of the reactants is

40cm^3 of CH4

40cm^3of H2

100cm^3 of H2O

Explanation:

Balanced reaction is

CH4 +H2+5/2O2______

CO2 +3H2O

Excess KOH at room temperature absorbs CO2 whose volume is given by 40cm^3 i.e the volume by which the solution decreases

So using Gay lussac combining ratio which states that gases combine in volumes that are in simple ratio to each other if gases.

Since CO2 in the equation is 1 mole

Means 1mole represent 40cm^3

So CH4:H2:O2 are in ratio of 1:1:5/2=(40:40:100)cm^3 respectively.

8 0

3 years ago

Decane (C10H22) is used in diesel. The combustion for decane follows the equation: 2 C10H22 + 31 O2 à 20 CO2 + 22 H2O. Calculate

creativ13 [48]

The mass of water produced is 792 grams by the combustion of 568 grams of decane.

Given:

Combustion of 568 grams of decane with 2979 grams of oxygen.

$2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O$

To find:

The mass of water produced by combustion of 568 grams of decane.

Solution:

Mass of decane = 568 g

Moles of decane :

= $\frac{568 g}{142 g/mol}=4 mol$

Mass of oxygen gas = 2976 g

Moles of oxygen gas:

= $\frac{2976 g}{32 g/mol}=93 mol$

$2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O$

According to reaction, 2 moles of decane reacts with 31 moles of oxygen, then 4 moles of decane will react with:

$=\frac{31}{2}\times 4mol=62\text{ mol of}O_2$

But according to the question, we have 93.0 moles of oxygen gas which is more than 62 moles of oxygen gas.

So, this means that oxygen gas is present in an excessive amount. Which simply means:

Oxygen gas is an excessive reagent.
Decane is a limiting reagent.
Decane being limiting reagent will be responsible for the amount of water produced after the reaction.

According to reaction, 22 moles of water is produced from 2 moles of decane, then 4 moles of decane will produce:

$=\frac{22}{2}\times 4mol=44\text{mol of }H_2O$

Mass of 44 moles of water ;

$=44mol\times 18g/mol=792g$

792 grams of water is produced by the combustion of 568 grams of decane.

Learn more about limiting reagent and excessive reagent here:

brainly.com/question/14225536?referrer=searchResults

brainly.com/question/7144022?referrer=searchResults

3 0

3 years ago

Hemos preparado la siguiente disolución 40 g de sal en 250 g de agua. Escribe:

Anestetic [448]

Calcula el Tanta por ciento en peso de soluto

5 0

4 years ago

How is the kinetic energy of the constituent particles of any matter affected by a change in temperature?

SpyIntel [72]

Answer:

Particles in all states of matter are in constant motion and this is very rapid at room temperature. A rise in temperature increases the kinetic energy and speed of particles; it does not weaken the forces between them. ... Individual particles in liquids and gases have no fixed positions and move chaotically.

Explanation:

5 0

3 years ago

A quantity of 0.0250 mol of a gas initially at 0.050 L and 19.0°C undergoes a constant-temperature expansion against a constant

KiRa [710]

Answer:

$V_2=2.995L\\\\W=248.5J$

Explanation:

Hello,

In this case, for us to compute the final volume we apply the Boyle's law that analyzes the pressure-volume temperature as an inversely proportional relationship:

$P_1V_1=P_2V_2$

So we solve for $V_2$ by firstly computing the initial pressure:

$P_1=\frac{nRT}{V_1}=\frac{0.025mol*0.082\frac{atm*L}{mol*K}*(19+273.15)K}{0.050L} =11.98atm$

$V_2=\frac{P_1V_1}{P_2}=\frac{11.98atm*0.050L}{0.200atm}\\ \\V_2=2.995L$

Finally, we can compute the work by using the following formula:

$W=nRTln(\frac{V_2}{V_1} )=0.025mol*8.314\frac{J}{mol*K}*(19.0+273.15)K*ln(\frac{2.995L}{0.050L}) \\\\W=248.5J$

Best regards.

4 0

3 years ago