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6) Hydrogen gas can be generated from the reaction between aluminum metal and hydrochloric acid:
2 Al(s) + 6 HCl(aq) + 2 AICI3, (aq) + 3 H2(g)
a. Suppose that 3.00 grams of Al are mixed with excess acid. If the hydrogen gas produced is directly collected
into a 850 mL glass flask at 24.0 °C, what is the pressure inside the flask (in atm)?
b. This hydrogen gas is then completely transferred from the flask to a balloon. To what volume (in L) will the
balloon inflate under STP conditions?
c. Suppose the balloon is released and rises up to an altitude where the temperature is 11.2 °C and the pressure is
438 mm Hg. What is the new volume of the balloon (in L)?

Respuesta :

pstnonsonjoku

Stoichiometry refers to the relationship between the moles of reactants and products.

This question must be solved using both stoichiometry and the gas laws

The reaction equation is;

2 Al(s) + 6 HCl(aq) --------> 2 AICI3, (aq) + 3 H2(g)

  • Using stoichiometry

Number of moles of Al = 3g/27g/mol = 0.11 moles

According to the reaction equation;

2 moles of Al yields 3 moles of H2

0.11 moles of Al yields 0.11 * 3/2 = 0.165 moles

  • Using the gas laws

From the ideal gas equation;

PV=nRT

P = ?

n= 0.165 moles

V = 0.85 L

T = 297 K

R = 0.082 atmLK-1mol-1

P= nRT/V

P = 0.165 * 0.082 * 297/0.85

P= 4.73 atm

  • Under STP conditions;

P1 = 4.73 atm

T1 = 297 K

V1 = 0.85 L

P2 = 1 atm

T2 =273 K

V2 =?

  • From the general gas equation;

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 =  4.73 * 0.85 * 273/1 * 297

V2 = 3.69 L

P1 = 760 mmHg

T1 = 273 K

V1 = 3.69

P2 = 438 mm Hg

T2 = 284.2 K

V2 =?

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 = 760 * 3.69 * 284.2/438 *273

V2 = 797010.48/119574

V2= 6.67 L

https://brainly.com/question/1190311

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