Answer:
The 2292 moles of CO are needed to react completely with 122 Kg of Fe₂O₃.
Explanation:
Given data:
Mass of Fe₂O₃ = 122 Kg ( 122×1000 = 122000 g)
Moles of CO = ?
Solution:
Chemical equation:
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Number of moles of Fe₂O₃:
Number of moles = mass/ molar mass
Number of moles = 122000 g /159.69 g/mol
Number of moles = 764 mol
Now we will compare the moles of Fe₂O₃ with CO.
Fe₂O₃ : CO
1 : 3
764 : 3×764 =2292 mol
The 2292 moles of CO are needed to react completely with 122 Kg of Fe₂O₃.
Answer:
Mg
Explanation:
The standard reduction potentials are
<u>E°/V
</u>
Au³⁺(aq ) + 3e⁻ ⟶ Au(s); 1.42
Hg²⁺(aq) + 2e⁻ ⟶ Hg(l); 0.85
Ag⁺(aq) + e⁻ ⟶ Ag(s); 0.80
Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34
Mg2+(aq) + 2e- ⟶ Mg(s); -2.38
The more negative the standard reduction potential, the stronger the metal is as a reducing agent.
Mg is the only metal with a standard reduction potential lower than that of Cu, so
Only Mg will react spontaneously with Cu²⁺.
The balanced equation for the reaction between Mg and HCl is as follows
Mg + 2HCl --> MgCl₂ + H₂
stoichiometry of HCl to H₂ is 2:1
number of HCl moles reacted - 0.400 mol/L x 0.100 L = 0.04 mol of HCl
since Mg is in excess HCl is the limiting reactant
number of H₂ moles formed - 0.04/2 = 0.02 mol of H₂
we can use ideal gas law equation to find the volume of H₂
PV = nRT
where
P - pressure - 1 atm x 101 325 Pa/atm = 101 325 Pa
V - volume
n - number of moles - 0.02 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 0 °C + 273 = 273 K
substituting these values in the equation
101 325 Pa x V = 0.02 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K
V = 448 x 10⁻⁶ m³
V = 448 mL
therefore answer is
c. 448 mL
Answer:
Step 1 should be convert atoms to moles (n). Step 2 should be convert moles (n) to mass (m).
Step 1
Use dimensional analysis to convert the number of atoms to moles.
1 mole atoms = 6.022 × 10²³ atoms
n(Ag) = 2.3 × 10²⁴ Ag atoms × (1 mol Ag/6.022 × 10²³ Ag atoms) = 3.8193 mol Ag
Step 2
Convert the moles of Ag to mass.
mass (m) = moles (n) × molar mass (M)
n(Ag) = 3.8193 mol Ag
M(Ag) = atomic weight on the periodic table in g/mol = 107.868 g Ag/mol Ag
m(Ag) = 3.8193 mol × 107.868 g/mol = 412 g Ag = 410 g Ag rounded to two significant figures
The mass of 2.3 × 10²⁴ Ag atoms is approximately 410 g.
Explanation:
