Question

Can any one solve this quiz plz​

Answer 1

1. It is stoichiometric.

2. O2 is the limiting reactant.

3. 9.0 g of C2H6 remain unreacted.

4. 17.6 g of CO2.

5. 85.2%.

Step-by-step explanation:

Hello there!

In this case, for the given chemical reaction:

`2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O`

We can see that:

1. It is stoichiometric and is balanced because the reactants yields the products according to the law of conservation of mass.

2. In this part, it is possible to calculate the moles of ethane by using its molar mass:

`n_(C_2H_6)=15g*(1molC_2H_6)/(30.08g) =0.50molC_2H_6`

And the moles of oxygen by knowing that one mole is contained in 22.4 L at STP:

`n_(O_2)=(1mol)/(22.4L) *15.68L=0.7molO_2`

Thus, by calculating the moles of carbon dioxide product by each reactant, we can identify the limiting one:

`n_(CO_2)^(by\ C_2H_6)=0.50molC_2H_6*(4molCO_2)/(2molC_2H_6) =1.0molCO_2\\\\n_(CO_2)^(by\ O_2)=0.70molO_2*(4molCO_2)/(7molO_2) =0.4molCO_2\\`

Thus, since oxygen yields the fewest moles of CO2 product, we infer it is the limiting reactant.

3. In this part, we calculate the mass of C2H6 that actually react first:

`m_(C_2H_6)^(reacted)=0.4molCO_2*(2molC_2H_6)/(4molCO_2)*(30.08gC_2H_6)/(1molC_2H_6) =6.0gC_2H_6`

Thus, the leftover of ethane (C2H6) as the excess reactant is:

`m_(C_2H_6 )^(leftover)=15g-6.0g=9.0g6.0C_2H_6`

4. Since 0.4 moles of carbon dioxide were produced, we use its molar mass to calculate the mass as its theoretical yield:

`m_(O_2)^(theoretical)=0.4molCO_2*(44gCO_2)/(1molCO_2)=17.6gCO_2`

5. Finally, the percent yield is gotten by dividing the actual yield by the theoretical one:

`Y=(15g)/(17.6)*100\%\\\\Y=85.2\%`

Best regards!