the work done in an open vessel at 300K , when 56g iron reacts with dilute HCL is
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atomic mass of Fe: 56 , of Cl = 37.5
Fe + 2 H Cl ==> Fe Cl2 + H2
1 mole 2 moles 1 mole 1 mole
When 56 g of Iron reacts with 2 moles of H Cl (assuming that there are at least 2 moles in the open vessel), one mole of Hydrogen gas comes out.
Initial volume of H2 gas = V1 = 0
Final volume of H2 gas: V2
P V2 = n R T = 1 * 8.314 * 300 = 2494.2 Joules
Work done = P (V2 - V1) = P * V2 = 2, 492.2 J
We can also substitute P = 1 atm = 1.013 * 10^5 Pa
and V = 22.400 Litres * 300°K / 273.15°K and then find PV. We get the same result.
Fe + 2 H Cl ==> Fe Cl2 + H2
1 mole 2 moles 1 mole 1 mole
When 56 g of Iron reacts with 2 moles of H Cl (assuming that there are at least 2 moles in the open vessel), one mole of Hydrogen gas comes out.
Initial volume of H2 gas = V1 = 0
Final volume of H2 gas: V2
P V2 = n R T = 1 * 8.314 * 300 = 2494.2 Joules
Work done = P (V2 - V1) = P * V2 = 2, 492.2 J
We can also substitute P = 1 atm = 1.013 * 10^5 Pa
and V = 22.400 Litres * 300°K / 273.15°K and then find PV. We get the same result.
kvnmurty:
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