Answer:
mass of sodium reacted is 184.1 g
Explanation:
mass Na = X = ?
∴ mass NaCl = 468 g
∴ mass Cl = 0.248 g
∴ molar mass NaCl = 58.44 g/mol
∴ atomic mass Cl = 35.453 a.m.u
∴ atomic mass Na = 22.989 a.m.u
⇒ moles Na = (X gNa)*(mol Na/22.989 g) = X/22.989 mol Na
⇒ mass NaCl = (X/22.989 mol Na)*(mol NaCl/mol Na)*(58.44 gNaCl/mol NaCl) = 468 g NaCl
clearing "X":
⇒ ((58.44)(X))/(22.989) = 468 g
⇒ X = 184.1 g = mass Na reacted
Answer:
0.533 mol O2
Explanation:
4 Fe3O4 + O2 -> 6 Fe2O3
1 mol O2 -> 6 mol Fe2O3
x -> 3.2 mol Fe2O3
x = (3.2 mol Fe2O3 * 1 mol O2)/ 6 mol Fe2O3
x= 0.533 mol O2
Energy may only be transformed from one sort to another. hope this helps :)
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C
The answer is 341.7 g.
(1) Calculate the molar mass (M) of CaCl2 which is the sum of atomic masses (A) of elements:
M(CaCl2) = A(Ca) + 2A(Cl)
A(Ca) = 40.1 g/mol
A(Cl) = 35.45 g/mol
M(CaCl2) = 40.1 + 2 * 35.45 = 40.1 + 70.9 = 111 g/mol
(2) Calculate in how many moles are 535 g:
M(CaCl2) = 111 g/mol
111g : 1mol = 535 g : xmol
x = 535 g * 1mol : 111g = 4.82 mol
(3) Calculate how many grams of Cl are in 4.82 mol:
A(Cl) = 35.45 g/mol
2A(Cl) = 2 * 35.45 = 70.9 g/mol
70.9 g : 1 mol = x : 4.82 mol
x = 70.9 g * 4.82 mol : 1 mol = 341.7 g
