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3 years ago

C2 H6 +02 equals CO2 + H2O what mass of water is produced from three moles of C2 H6?

Chemistry

1 answer:

yulyashka [42]3 years ago

8 0

Answer:

162g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2C2H6 + 7O2 —> 4CO2 + 6H2O

Next, we shall determine the number of mole of water, H2O produced by the reaction of 3 moles of C2H6.

This can be obtained as follow:

From the balanced equation above,

2 moles of C2H6 reacted to produce 6 moles of H2O.

Therefore, 3 moles of C2H6 will react to produce = (3 x 6)/2 = 9 moles of H2O.

Therefore, 9 moles of H2O is produced from the reaction.

Finally, we shall convert 9 moles of H2O to grams.

This can be done as shown below:

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mole of H2O = 9 moles

Mass of H2O =..?

Mole = mass / molar mass

9 = mass of H2O /18

Cross multiply

Mass of H2O = 9 x 18

Mass of H2O = 162g

Therefore, 162g of H2O were produced from 3 moles of C2H6.

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In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class t

solmaris [256]

Answer:

4.81 moles

Explanation:

The total pressure of the gas = Pressure at which gauge reads zero + pressure read by it.

Pressure at which gauge reads zero = 14.7 psi

Pressure read by the gauge = 988 psi

Total pressure = 14.7 + 988 psi = 1002.7 psi

Also, P (psi) = P (atm) / 14.696

Pressure = 1002.7 / 14.696 = 68.2297 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25 + 273.15) K = 298.15 K

Volume = 1.50 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

68.2297 atm × 1.5 L = n × 0.0821 L.atm/K.mol × 298.15 K

⇒n = 4.81 moles

4 0

3 years ago

What reaction is depicted by the given equation: au3 3e− au

mrs_skeptik [129]

This is a reduction half-reaction. Initially, gold exists as $Au^{3+}$ , and after the reaction, it has been converted to elemental gold. To make this possible, three electrons must be added to $Au^{3+}$ . This fits within the definition of reduction reactions, where the reactant gains electrons in the reaction.