The overall cell reaction occurring in an alkaline battery is:
Zn(s) + MnO2(s) + H2O(l) ---> ZnO(s) + Mn(OH)2(s)
a) How many moles of electrons flow per mole of reaction?...answer: 2. Mol e-Note:
b) If 6.09g of zinc is oxidized, how many grams of manganese dioxide and of water are consumed?
Grams MNnO2 answer : 8.092 g
grams H2O answer: 1.676 g
c) What is the total mass of reactants consumed in part b?
answer: 15.858 g
d) How many coulombs are produced in part (b)?.....I need this answer in scientific notation.
Thank you...
I figured it out...it's 1.79 x 10^4
To find the number of moles of electrons flowing per mole of reaction, we need to balance the chemical equation. From the balanced equation:
2 moles of electrons flow per mole of reaction.
b) To find the grams of manganese dioxide and water consumed when 6.09 g of zinc is oxidized, we can use the mole ratio from the balanced equation:
From the balanced equation, the mole ratio between zinc and manganese dioxide is 1:1, and the mole ratio between zinc and water is 1:1.
So, the grams of manganese dioxide consumed = 6.09 g
The grams of water consumed = 6.09 g
c) The total mass of reactants consumed in part b is the sum of the masses of manganese dioxide and water consumed:
Total mass of reactants consumed = 6.09 g (manganese dioxide) + 6.09 g (water) = 12.18 g
d) To find the number of coulombs produced in part b, we need to convert the mass of zinc oxidized to moles using the molar mass of zinc, and then use Faraday's constant (96485 C/mol e-) to convert moles of electrons to coulombs.
The molar mass of zinc (Zn) is 65.38 g/mol. So, the number of moles of zinc oxidized is:
Number of moles of zinc = 6.09 g / 65.38 g/mol = 0.0932 mol
Since 2 moles of electrons flow per mole of reaction, the number of moles of electrons involved is also 0.0932 mol.
The number of coulombs produced can be calculated as:
Number of coulombs = 0.0932 mol e- × (96485 C/mol e-) = 8986 C
In scientific notation, this is 8.986 × 10^3 C.
To find the answer to part (d), we first need to calculate the number of moles of zinc oxidized in part (b).
The molar mass of zinc (Zn) is approximately 65.38 g/mol.
Therefore, the number of moles of zinc can be calculated as follows:
Number of moles of zinc = mass of zinc / molar mass of zinc
= 6.09 g / 65.38 g/mol
= 0.0932 mol
Now, we know from the balanced equation that 2 moles of electrons flow per mole of reaction. Therefore, 2 moles of electrons flow for every 1 mole of zinc oxidized.
So, the number of moles of electrons produced can be calculated as:
Number of moles of electrons = 2 moles of electrons/mol of zinc * number of moles of zinc
= 2 * 0.0932 mol
= 0.1864 mol
Next, we need to convert the number of moles of electrons to coulombs.
1 mole of electrons is equal to 6.022 x 10^23 electrons.
1 coulomb is equal to 6.242 x 10^18 electrons.
Therefore, the number of coulombs produced can be calculated as:
Number of coulombs = number of moles of electrons * (6.022 x 10^23 electrons / 1 mole of electrons) * (1 coulomb / 6.242 x 10^18 electrons)
= 0.1864 mol * (6.022 x 10^23 electrons / 1 mol) * (1 coulomb / 6.242 x 10^18 electrons)
Now, let's simplify this expression:
Number of coulombs = 0.1864 * (6.022 / 6.242) * (10^23 / 10^18)
≈ 0.1791 * 10^5
≈ 1.791 x 10^4 coulombs
Therefore, the number of coulombs produced in part (b) is approximately 1.791 x 10^4 coulombs.