To determine the moles of oxygen required for the complete combustion of 1 mole of propane (C\(_3\)H\(_8\)), we must first write the balanced chemical equation for the combustion process. The combustion of propane is represented by:
C\(_3\)H\(_8\) + 5O\(_2\) → 3CO\(_2\) + 4H\(_2\)O
This equation indicates that 1 mole of propane (C\(_3\)H\(_8\)) reacts with 5 moles of oxygen (O\(_2\)) to produce 3 moles of carbon dioxide (CO\(_2\)) and 4 moles of water (H\(_2\)O).
Therefore, 5 moles of oxygen are required to completely combust 1 mole of propane.
| Substance | Moles in Reaction |
|---|---|
| C3H8 | 1 |
| O2 | 5 |
| CO2 | 3 |
| H2O | 4 |
Fortification of food with iron is done using $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$. The mass in grams of the $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$ required to achieve 12 ppm of iron in 150 kg of wheat is _______ (Nearest integer).} (Given : Molar mass of $\mathrm{Fe}, \mathrm{S}$ and O respectively are 56,32 and $16 \mathrm{~g} \mathrm{~mol}^{-1}$ )
For the reaction \( A + B \to C \), the rate law is found to be \( \text{rate} = k[A]^2[B] \). If the concentration of \( A \) is doubled and \( B \) is halved, by what factor does the rate change?