Propane burns in air according to the equation: C_3H_8(g) + 5O_2 (g) rightarrow

Question

Propane burns in air according to the equation: C_3H_8(g) + 5O_2 (g) rightarrow 3CO_2 (g) + 4H_2O (g) What volume of CO_2 would be formed if 5.00 L of propane burns, assuming that all of the gases are under the same conditions? A) 5.00 L B) 15.0 L C) 3.00 L D) 4.00 L E) not enough information What volume of H_2 would be collected at 22.0 degree C and a pressure 713 torr if 2.65 g of zinc react according to the equation: Zn(s) + 2HCl (aq) rightarrow ZnCl_2 (aq) + H_2 (g) A) 0.908 L B) 0.0405 L C) 1.04 L D) 2.08 L E) 0.523 L

Explanation / Answer

33)

From given reaction,

mol of CO2 formed = 3*mol of C3H8 burnt

Since all gases are under standard condition:

volume of CO2 formed = 3*volume of C3H8 burnt

= 3*5.00 L

= 15.0 L

Answer: B

34)

molar mass of Zn = 65.4 g/mol

number of mol of Zn = mass / molar mass

= 2.65 g / 65.4 g/mol

= 0.0405 mol

from given reaction,

mol of H2 formed = mol of Zn reacted

= 0.0405 mol

so we have:

n = 0.0405 mol

P =713 torr = 713/760 atm = 0.938 atm

T = 22.0 oC = (22.0 + 273) K = 295 K

use:

P*V = n*R*T

0.938 atm * V = 0.0405 mol * 0.0821 atm.L/mol.K * 295 K

V = 1.04 L

Answer: C

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