How much energy is obtained if 11.2g of hydrogen gas are allowed to react with excess fluorine?
H2 (g) + F2 (g)--- 2HF (g) + 124kcal
124 kcal x (11.2 g HF/molar mass H2) = ?
To calculate the amount of energy obtained when 11.2g of hydrogen gas reacts with excess fluorine, we need to use the given balanced chemical equation and the energy change of the reaction.
The balanced chemical equation is:
H2 (g) + F2 (g) → 2HF (g)
The energy change of the reaction is given as 124 kcal.
First, we need to convert the mass of hydrogen gas (H2) to moles. To do this, we use the molar mass of hydrogen, which is approximately 2 g/mol.
Number of moles of H2 = Mass of H2 / Molar mass of H2
Number of moles of H2 = 11.2 g / 2 g/mol
Number of moles of H2 = 5.6 mol
From the balanced chemical equation, we can see that 1 mole of H2 reacts with 1 mole of F2 to produce 2 moles of HF. Therefore, the number of moles of F2 required for the reaction is also 5.6 mol.
Now, since there is an excess of fluorine, all the hydrogen gas will be used up. Therefore, the reaction will produce 5.6 moles of HF.
Next, we need to calculate the energy obtained from the reaction. The given energy change is 124 kcal for 1 mole of reaction. Since there are 2 moles of HF produced in the reaction, we can calculate the energy obtained using the following:
Energy obtained = Energy change per mole of reaction × Number of moles of reaction
Energy obtained = 124 kcal/mol × 5.6 mol
Energy obtained = 694.4 kcal
Therefore, if 11.2g of hydrogen gas is allowed to react with excess fluorine, the amount of energy obtained is 694.4 kcal.