answer is one first one for one electron.
1) The equation is already balanced;
Left side: 1C, 2H and 1O
Right side: 1C, 1O, 2H
2) Starting substances: water and carbon
There are two starting substances also called reactants. H2O which is a compound and C which is an element.
3) Ending substance asked for: H2.
This is hydrogen gas.
Stoichiometry:
1 mol of C produces 1 mol of H2
12.01 g of C productes 2.016 grams of H2
Ratio: 2.016 g of H2 / 12.01 g of C = X g of H2 / 34 g of C =>
x = 34 g of C* 2.016 g of H2 / 12.01 g of C= 5.707 g of H2
2MnO4-(aq) + 16H+ + 10Cl-(aq) ⇔ 2Mn2+(aq) + 8H2O + 5Cl2 (g)
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
Mno4- (aq) + Cl- (aq) → Mn2+ + Cl2 (g)
Required
Half reaction
Solution
1. Add coefficient(equalizing atoms in reaction)
2. Adding H₂O on the O-deficient side.
3. Adding H⁺ on the H-deficient side.
4. Adding e⁻
5. Equalizing the number of electrons and sum the all the half-reaction
1.MnO₄⁻(aq) = Mn²⁺(aq) reduction
2.MnO4(aq) = Mn2+(aq) + 4H2O
3. MnO4-(aq) + 8H+ = Mn2+(aq) + 4H2O
4. MnO4-(aq) + 8H+ + 5e- = Mn2+(aq) + 4H2O
Cl⁻(aq) = Cl₂(g) oxidation
1. 2Cl-(aq) = Cl2 (g)
2-3 none
4. 2Cl-(aq) = Cl2 (g) + 2e-
5.
MnO4-(aq) + 8H+ + 5e- = Mn2+(aq) + 4H2O x2
2Cl-(aq) = Cl2 (g) + 2e- x5
2MnO4-(aq) + 16H+ + 10e- = 2Mn2+(aq) + 8H2O
10Cl-(aq) = 5Cl2 (g) + 10e-
<em>2MnO4-(aq) + 16H+ + 10Cl-(aq) ⇔ 2Mn2+(aq) + 8H2O + 5Cl2 (g)</em>
The answer is -0.22 for the total impact of velocity 1.1 seconds -5.0=-
1) first, we have to convert the grams to moles of AuCl3 using the molar mass of the molecule.
molar mass of AuCl3= 197 + (35.5 x 3)= 303.5 g/mol
73.4 g (1 mol AuCl3/ 303.5 g)= 0.242 moles
2) now. let's convert moles of AuCl3 to moles of chlorine gas (Cl2) using the mole-mole ratio
0.242 mol AuCl3 (3 mol Cl2/ 2 mol AuCl3)= 0.363 mol Cl2
3) finally, we convert moles to grams using the molar mass of Cl2.
molar mass of Cl2 = 35.5 x 2= 71.0 g/mol
0.363 mol Cl2 ( 71.0 g/ 1 mol)= 25.8 grams