What volume of so2 is produced at 325 k and 1.35 atm when 15.0 grams of hcl reacts with excess k2so3?

1 answer:

3 0

The volume of SO2 produced at 325k is calculated as below

calculate the moles of SO2 produced which is calculated as follows

write the reacting equation
K2SO3 +2 HCl = 2KCl +H2O+ SO2

find the moles of HCl used
=mass/molar mass = 15g/ 36.5 g/mol =0.411 moles

by use of mole ratio between HCl to SO2 which is 2:1 the moles of SO2 is therefore = 0.411 /2 =0.206 moles of SO2

use the idea gas equation to calculate the volume SO2
that is V=nRT/P
where n=0.206 moles
R(gas constant) = 0.082 L.atm/ mol.k
T=325 K
P=1.35 atm

V=(0.206 moles x 0.082 L.atm/mol.k x325 k)/1.35 atm = 4.07 L of SO2

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Balloon A started with a volume of 1.76 L. The temperature of the room the balloon was in was 295 K. The balloon was heated to a

valentina_108 [34]

Answer:

V₂ = 1.5 L

Explanation:

Given data:

Initial volume of balloon = 1.76 L

Initial temperature = 295 K

Final temperature = 253.15 K

Final volume = ?

Solution:

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 1.76 L ×253.15 K / 295 K

V₂ = 445.54 L.K /295 K

V₂ = 1.5 L

4 0

3 years ago

Calculate the approximate volume of a 0.6000mol sample of gas at 288.15K and a pressure of 1.10atm.

11111nata11111 [884]

Answer:

The volume of the sample of the gas is found to be 12.90 L.

Explanation:

Given pressure of the gas = P = 1.10 atm

Number of moles of gas = n = 0.6000 mole

Temperature = T = 288.15 K

Assuming the volume of the gas to be V liters

The ideal gas equation is shown below

$\textrm{PV} =\textrm{nRT} \\1.10 \textrm{ atm}\times V \textrm{ L} = 0.6000 \textrm{ mole}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 288.5\textrm{K} \\\textrm{V} = 12.90 \textrm{ L}$