This problem is pointing out acetylene gas undergoes combustion, so it asks for the equation, limiting reactant, excess, moles of target product and how much of the excess reactant is left.
Summary of answers:
- [tex]2CH_2+3O_2\rightarrow 2CO_2+2H_2O[/tex]
- Limiting reactant: acetylene.
- Excess reactant: oxygen.
- moles of target product produced: 3.5 moles.
- moles of excess reagent remaining: 0.25 moles.
Uses of stoichiometry:
In chemistry, when two substances react to produce specific products, one can use the law of conservation of mass to correctly write the chemical equation standing for the chemical change. Thus, for acetylene, when it undergoes combustion, we would have:
[tex]2CH_2+3O_2\rightarrow 2CO_2+2H_2O[/tex]
In such a way, one can identify the limiting and excess reactants and the moles of target product, carbon dioxide whose molar mass is 44 g/mole, with the following calculations based on the mole ratios derived from the chemical equation:
[tex]3.5molCH_2*\frac{2molCO_2}{2molCH_2}=3.5molCO_2\\ \\5.5molO_2*\frac{2molCO_2}{3molO_2}=3.7molCO_2[/tex]
In such a way, we conclude that acetylene is the limiting reactant because it produces the lowest moles of CO2 and thus, oxygen is in excess. Moreover, the produced moles of the target product are 3.5 mol.
Finally, we can calculate the consumed oxygen with the following mole ratio and the leftover of oxygen (excess):
[tex]3.5molCH_2*\frac{3molO_2}{2molCH_2} =5.25molO_2[/tex]
[tex]mol _{O_2}^{left}=5.5mol-5.25mol=0.25mol[/tex]
Learn more about stoichiometry: brainly.com/question/9743981
