We know that there are 100 cm in 1 m, so we need to add that into the conversion:
![(\frac{0.00082 m}{1})*(\frac{1m}{100cm})*(\frac{5.06cm}{1})=0.00004149 m^{2}](https://tex.z-dn.net/?f=%20%28%5Cfrac%7B0.00082%20m%7D%7B1%7D%29%2A%28%5Cfrac%7B1m%7D%7B100cm%7D%29%2A%28%5Cfrac%7B5.06cm%7D%7B1%7D%29%3D0.00004149%20m%5E%7B2%7D%20%20%20%20)
Or, in scientific notation:
![4.15x10^{-5}m^{2}](https://tex.z-dn.net/?f=%204.15x10%5E%7B-5%7Dm%5E%7B2%7D%20)
Answer: The limiting reactants for the reaction is, Mg
Explanation : Given,
Mass of
= 10.1 g
Mass of
= 10.5 g
Molar mass of
= 24 g/mol
Molar mass of
= 32 g/mol
First we have to calculate the moles of
and
.
![\text{Moles of }Mg=\frac{\text{Given mass }Mg}{\text{Molar mass }Mg}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DMg%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%20%7DMg%7D%7B%5Ctext%7BMolar%20mass%20%7DMg%7D)
![\text{Moles of }Mg=\frac{10.1g}{24g/mol}=0.421mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DMg%3D%5Cfrac%7B10.1g%7D%7B24g%2Fmol%7D%3D0.421mol)
and,
![\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DO_2%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%20%7DO_2%7D%7B%5Ctext%7BMolar%20mass%20%7DO_2%7D)
![\text{Moles of }O_2=\frac{10.5g}{32g/mol}=0.328mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DO_2%3D%5Cfrac%7B10.5g%7D%7B32g%2Fmol%7D%3D0.328mol)
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
![2Mg(s)+O_2(g)\rightarrow 2MgO(s)](https://tex.z-dn.net/?f=2Mg%28s%29%2BO_2%28g%29%5Crightarrow%202MgO%28s%29)
From the balanced reaction we conclude that
As, 2 mole of
react with 1 mole of ![O_2](https://tex.z-dn.net/?f=O_2)
So, 0.421 moles of
react with
moles of ![O_2](https://tex.z-dn.net/?f=O_2)
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Hence, the limiting reactants for the reaction is, Mg
Answer:
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%.
Explanation:
Hello there!
In this case, for the given chemical reaction:
![2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O](https://tex.z-dn.net/?f=2C_2H_6%2B7O_2%5Crightarrow%204CO_2%2B6H_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*\frac{1molC_2H_6}{30.08g} =0.50molC_2H_6](https://tex.z-dn.net/?f=n_%7BC_2H_6%7D%3D15g%2A%5Cfrac%7B1molC_2H_6%7D%7B30.08g%7D%20%3D0.50molC_2H_6)
And the moles of oxygen by knowing that one mole is contained in 22.4 L at STP:
![n_{O_2}=\frac{1mol}{22.4L} *15.68L=0.7molO_2](https://tex.z-dn.net/?f=n_%7BO_2%7D%3D%5Cfrac%7B1mol%7D%7B22.4L%7D%20%2A15.68L%3D0.7molO_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*\frac{4molCO_2}{2molC_2H_6} =1.0molCO_2\\\\n_{CO_2}^{by\ O_2}=0.70molO_2*\frac{4molCO_2}{7molO_2} =0.4molCO_2\\](https://tex.z-dn.net/?f=n_%7BCO_2%7D%5E%7Bby%5C%20C_2H_6%7D%3D0.50molC_2H_6%2A%5Cfrac%7B4molCO_2%7D%7B2molC_2H_6%7D%20%3D1.0molCO_2%5C%5C%5C%5Cn_%7BCO_2%7D%5E%7Bby%5C%20O_2%7D%3D0.70molO_2%2A%5Cfrac%7B4molCO_2%7D%7B7molO_2%7D%20%3D0.4molCO_2%5C%5C)
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*\frac{2molC_2H_6}{4molCO_2}*\frac{30.08gC_2H_6}{1molC_2H_6} =6.0gC_2H_6](https://tex.z-dn.net/?f=m_%7BC_2H_6%7D%5E%7Breacted%7D%3D0.4molCO_2%2A%5Cfrac%7B2molC_2H_6%7D%7B4molCO_2%7D%2A%5Cfrac%7B30.08gC_2H_6%7D%7B1molC_2H_6%7D%20%3D6.0gC_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](https://tex.z-dn.net/?f=m_%7BC_2H_6%20%7D%5E%7Bleftover%7D%3D15g-6.0g%3D9.0g6.0C_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*\frac{44gCO_2}{1molCO_2}=17.6gCO_2](https://tex.z-dn.net/?f=m_%7BO_2%7D%5E%7Btheoretical%7D%3D0.4molCO_2%2A%5Cfrac%7B44gCO_2%7D%7B1molCO_2%7D%3D17.6gCO_2)
5. Finally, the percent yield is gotten by dividing the actual yield by the theoretical one:
![Y=\frac{15g}{17.6}*100\%\\\\Y=85.2\%](https://tex.z-dn.net/?f=Y%3D%5Cfrac%7B15g%7D%7B17.6%7D%2A100%5C%25%5C%5C%5C%5CY%3D85.2%5C%25)
Best regards!
Answer:
Br2+ 2H2O + SO2= 2HBr + H2SO4
Let's Compare the left side of the equation to the right side of the equation.
Left: Br= 2, H= 2, S= 1, O = 1+2
Right: Br=1, H= 1+2, S=1, O= 4
We can see that only S is balanced and not the other 3 elements.
I'll try to make each element balance.
For Br; I'll multiply by 2 on the left to make it equal to the right.
For H; Since the 2 for Br on the right affected also H, that H ( for HBr) Already has a 2, but then it adds with the other H2( for H2SO4) to give a total of 4 H on the right side. But then there's only 2 H on the left. so we multiply that 2 by a 2 ( which is written infront of the H2O to give a total of 4 H on the left side.
For O; Because of the 2 infront of the H2O, it affects the O in H2O..so now we have 2 O plus the 2 O ( in SO2) to give a total of 4 O which is equal to the right side.