Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol
Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol
Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
= (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
= 176.15 kJ - 84.78 kJ = 91.38 kJ
Answer:
59.8%
Explanation:
First find the Mr of manganese (III) nitrate.
Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>
Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:
Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>
Now we can find percentage composition / percentage by mass of oxygen.
% composition = 
× 100
% composition = 
× 100 = <u>59.776%</u>
∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).
Answer:
39.2 L at STP
Explanation:
Convert the grams to moles first by dividing 56.0 by the molar mass of O2 (32.0) then convert to volume by multiplying by 22.4.
= 39.2 L
The product of the complete combustion of any fuel (in this case, acetylene) are indeed water and carbon dioxide.
Balancing the combustion reaction,
C2H2 +(5/2) O2 --> 2CO2 + H2O
The number of moles of C2H2 will be,
(12 g) x (1 mole/26 g) = 6/13 mole
Then, the number of moles of O2 is,
(12 g) x (1 mole/32 g) = 3/8 mole
Therefore the limiting reaction is the O2. Getting the amount of CO2 and H2O produced from balancing,
CO2 = (3/8 moles) x (2 moles CO2/ 5/2 mole O2)(44 g/ 1 mole) = 52.8 g
H2O = (3/8 moles) x (1 mole / 5/2 mole O2)(18 g / 1 mole) = 2.7 g
Answer:
В. No, because the mass of the reactants is less than the mass of the products.
Explanation:
Chemical equation:
NaBr + Cl₂ → 2NaCl + Br₂
The given equation is not balanced because number of moles of sodium and bromine atoms are less on reactant side while more on the product side.
There are one mole of sodium and one mole of bromine atom on left side of equation while on right side there are 2 moles of bromine and 2 moles of sodium atom are present. The number of moles of chlorine atoms are balanced.
Balanced chemical equation:
2NaBr + Cl₂ → 2NaCl + Br₂
Now equation is balanced. Number of moles of sodium , chlorine and bromine atoms are equal on both side.
