Respuesta :
What is Yield?
In chemistry, yield, also known as reaction yield, is a percentage measure of the number of moles of a product formed in relation to the reactant consumed in a chemical reaction. One of the most important factors that scientists must consider in organic and inorganic chemical synthesis processes is yield. The terms "yield", "conversion," and "selectivity" are used in chemical reaction engineering to describe ratios of how much of a reactant was consumed (conversion), how much desired product was formed (yield), and how much undesired product was formed (selectivity), represented as X, Y, and S.
Mole of C = 0.100mol
Mass of [tex]O_2[/tex] = 8.00g
Calculation
First, we translate the statement into the following chemical equation. The question is balance so now proceed to the calculation
[tex]\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)[/tex]
We Multiply the molar ratio between C and [tex]\mathrm{H}_{2} \mathrm{~S}[/tex]. Based on the chemical equation the molar ratio is [tex]1 \mathrm{~mol} \mathrm{C/1} \mathrm{mol} \mathrm{CO}_{2}[/tex]
Moles of CO[tex]_2[/tex] formed [tex]-0.100 \mathrm{~mol} \mathrm{C} \times \frac{1 \mathrm{~mol} \mathrm{CO}}{1 \mathrm{~mol} \mathrm{C}}-0.100 \mathrm{~mol} \mathrm{CO}_{2}[/tex]
For oxygen, we multiply the given mass of [tex]O_2[/tex] by the reciprocal of its molar mass. The molar mass of [tex]O_2\\[/tex] is 32.00g/mol. Then, we multiply the molar ratio between [tex]O_2 {and CO_2[/tex]. Based on the chemical equation, the molar ratio is [tex]1 mol CO_2/1 mole O_2[/tex].
Moles of [tex]CO_2[/tex] formed = [tex]8.00 \mathrm{~g} \mathrm{O}_{2} \times \frac{1 \mathrm{~mol} \mathrm{O}_{2}}{32.00 \mathrm{~g} \mathrm{O}} \times \frac{1 \mathrm{~mol} \mathrm{CO}}{1 \mathrm{~mol} \mathrm{C}_{2}}=0.250 \mathrm{~mol}[/tex][tex]CO_2[/tex]
We can see that the limiting reactant is C and the excess reactant is [tex]O_2[/tex]since the given moles (assuming complete reaction) of C forms lesser product than oxygen does. We now multiply the molar mass of [tex]CO_2[/tex] to the calculated number of moles. The molar mass of [tex]CO_2[/tex] is 44.01 g/mol
[tex]\text { Mass of } \mathrm{CO}_{2} \text { formed }=0.100 \mathrm{~mol} \mathrm{CO}_{2} \times \frac{14.01 \mathrm{~g} \mathrm{CO}_{2}}{1 \mathrm{~mol} \mathrm{CO}}=4.401 \text { or } 4.40 \mathrm{~g} \mathrm{CO_2}[/tex]
We multiply the number of moles of C by the molar ratio between C and [tex]O_2[/tex] which is 1 mol O2 /1 mol C to get the number of moles of [tex]O_2[/tex] reacted. We the multiply the molar mass.
[tex]\text { Mass of } \mathrm{O}_{2} \text { reacted }=0.100 \mathrm{~mol} \mathrm{C} \times \frac{1 \mathrm{~mol} \mathrm{O}_{2}}{1 \mathrm{~mol} \mathrm{C}} \times \frac{32.00 \mathrm{~g} \mathrm{O}_{2}}{1 \mathrm{~mol} \mathrm{O}}=3.20g \mathrm{O_2}[/tex]
We subtract the the mass of [tex]O_2[/tex] reacted from the given mass of [tex]O_2[/tex]
Mass of excess [tex]O_2 = 800g - 3.20g = 4.80g O_2[/tex]
Final Answer :
The excess reaction is [tex]O_2[/tex]
Mass of [tex]CO_2[/tex] formed - 4.40g [tex]CO_2[/tex]
Mass of excess [tex]O_2[/tex] = 4.80g [tex]O_2[/tex]
To learn more about Yield
brainly.com/question/25996347
#SPJ4
