The answer is: 17.0 g
I hope it help!
To calculate percent composition, you first need to find the molar mass of C (carbon), H (hydrogen) and O (oxygen).
C is 12.01
H is 1.00
O is 16
Then multiply each by the number of atoms of each element in the formula (the number that comes after each element in the equation for example C6 means 6 carbon atoms.
C: 12.01 x 6= 72.06
H: 1x12= 12
O: 16x6= 96
Then add them up.
72.06+ 12+ 96= 180.06
Now find the percent composition of carbon.
72.06/ 180.06 x 100= 40.01%
So the answer is C 40%.
Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100
∴ if theoretical yield is 26 g, but only 22.0 is recovered from the reaction,
then Percentage Yield = (22 g ÷ 26 g) × 100
= 84.6 %
Answer:
20.4 grams Zn
Explanation:
To find the mass, you first need to find the moles. This can be found using the Ideal Gas Law equation:
PV = nRT
In this equation,
-----> P = pressure (atm)
-----> V = volume (L)
-----> n = moles
-----> R = Ideal Gas Constant (0.08206 atm*L/mol*K)
-----> T = temperature (K)
Before you can plug the values into the equation, you need to convert Celsius to Kelvin.
P = 0.980 atm R = 0.08206 atm*L/mol*K
V = 7.80 L T = 25.0 °C + 273.15 = 298.15 K
n = ? moles
PV = nRT
(0.980 atm)(7.80 L) = n(0.08206 atm*L/mol*K)(298.15 K)
7.644 = n(24.466)
0.312 moles = n
Now that you have the number of moles, you can convert it to grams using the atomic mass of zinc. The final answer should have 3 sig figs to match the sig figs in the given values.
Atomic Mass (Zn): 65.380 g/mol
0.312 moles Zn 65.380 grams
------------------------- x ------------------------- = 20.4 grams Zn
1 mole
