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RoseWind [281]
3 years ago
5

`You have to be careful about pouring drano down your pipes since it is mainly hydrochloric acid--you can't do it if they are ma

de of aluminum because it will dissolve them! You can't acid clean your aluminum auto parts for the same reasons: 2 Al + 6 HCl → 2 AlCl3 + 3H2. If you were able to dissolve 10.0 mol of Al, how many many moles of hydrogen gas could you make? NOTE: All numbers located immediately after elemental symbols should be considered subscripts.
A. 3.00 moles
B. 6.00 moles
C. 1.50 moles
D. 15.0 moles
E. none of the above
Chemistry
1 answer:
notsponge [240]3 years ago
8 0

<u>Answer:</u> The number of moles of hydrogen gas that will be produced is 15.0 moles.

<u>Explanation:</u>

We are given:

Moles of aluminum needed to be dissolved = 10.0 moles

For the given chemical reaction:

2Al+6HCl\rightarrow 2AlCl_3+3H_2

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, 10.0 moles of aluminium will produce = \frac{3}{2}\times 10.0=15.0mol of hydrogen gas

Hence, the number of moles of hydrogen gas that will be produced is 15.0 moles.

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A propane stove burned 470 grams propane and produced 625 grams of water (this is the actual yield) C3H8 +5O2=3CO2+4H20. What wa
Liula [17]

Answer:

81.3%

Explanation:

Step 1:

The balanced equation for the reaction:

This is shown below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Step 2:

Data obtained from the question. This includes:

Mass of propane (C3H8) = 470 g

Actual yield of water (H2O) = 625 g

Percentage yield of water (H2O) =?

Step 3:

Determination of the mass of propane (C3H8) burned and the mass of water (H2O) produce from the balanced equation. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

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

Mass of H2O from the balanced equation = 4 x 18 = 72g

From the balanced equation above,

44g of C3H8 was burned and 72g of H2O was produced.

Step 4:

Determination of the theoretical yield of H2O. This is illustrated below:

From the balanced equation above,

44g of C3H8 produced 72g of H2O.

Therefore, 470g of C3H8 will produce = (470x72)/44 = 769.09g of H2O.

Therefore, the theoretical yield of H2O is 769.09g

Step 5:

Determination of the percentage yield of water (H2O). This is illustrated below:

Actual yield of water (H2O) = 625g

theoretical yield of H2O = 769.09g

Percentage yield of water (H2O) =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 625/769.09 x100

Percentage yield = 81.3%

Therefore, the percentage yield of water (H2O) is 81.3%

4 0
4 years ago
Calculate the heat change involved when 2.00 L of water is heated from 20.0/C to 99.7/C in
IgorC [24]

Answer:

669.48 kJ

Explanation:

According to the question, we are required to determine the heat change involved.

We know that, heat change is given by the formula;

Heat change = Mass × change in temperature × Specific heat

In this case;

Change in temperature = Final temp - initial temp

                                       = 99.7°C - 20°C

                                       = 79.7° C

Mass of water is 2000 g ( 2000 mL × 1 g/mL)

Specific heat of water is 4.2 J/g°C

Therefore;

Heat change = 2000 g × 79.7 °C × 4.2 J/g°C

                      = 669,480 joules

But, 1 kJ = 1000 J

Therefore, heat change is 669.48 kJ

7 0
3 years ago
One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu
Deffense [45]

Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

7 0
3 years ago
As an object falls freely near the earths surface , the loss in gravitational potential energy of the object is equal to its wha
Nitella [24]
Increase in kinetic energy as well as energy loss to the surroundings in the form of heat ( negligible)
4 0
3 years ago
Read 2 more answers
Is a burning log an exothermic or endothermic event if the log is the system?
denpristay [2]
It would be endothermic because the log is in the system.
6 0
3 years ago
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