<u>Answer:</u> The mass of decane produced is ![1.743\times 10^2g](https://tex.z-dn.net/?f=1.743%5Ctimes%2010%5E2g)
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
......(1)
Mass of hydrogen gas = 2.45 g
Molar mass of hydrogen gas = 2 g/mol
Putting values in equation 1:, we get:
![\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DH_2%3D%5Cfrac%7B2.45g%7D%7B2g%2Fmol%7D%3D1.225mol)
The chemical equation for the hydrogenation of decene follows:
![C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)](https://tex.z-dn.net/?f=C_%7B10%7DH_%7B20%7D%28l%29%2BH_2%28g%29%5Crightarrow%20C_%7B10%7DH_%7B22%7D%28s%29)
As, decene is present in excess. So, it is considered as an excess reagent.
Thus, hydrogen gas is a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
1 mole of hydrogen gas produces 1 mole of decane.
So, 1.225 moles of hydrogen gas will produce =
of decane
Now, calculating the mass of decane by using equation 1, we get:
Moles of decane = 1.225 mol
Molar mass of decane = 142.30 g/mol
Putting values in equation 1, we get:
![1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g](https://tex.z-dn.net/?f=1.225mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20decane%7D%7D%7B142.30g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20carbon%20dioxide%7D%3D%281.225mol%5Ctimes%20142.30g%2Fmol%29%3D174.3g%3D1.743%5Ctimes%2010%5E2g)
Hence, the mass of decane produced is ![1.743\times 10^2g](https://tex.z-dn.net/?f=1.743%5Ctimes%2010%5E2g)