Let's go through each of the problems step by step.
47. Decomposition of Potassium Nitrate
The unbalanced reaction is: \[ \text{KNO}_3 \rightarrow \text{KNO}_2 + \text{O}_2 \]
To balance it:
- Each KNO3 produces 1 KNO2 and half an O2 molecule. Therefore, we need to balance the O2 on the right side.
- Ultimately, we can balance it using coefficients:
\[ 2 \text{KNO}_3 \rightarrow 2 \text{KNO}_2 + \text{O}_2 \]
48. Reaction of Iron(III) Oxide with Carbon
The unbalanced reaction is: \[ \text{Fe}_2\text{O}_3 + \text{C} \rightarrow \text{Fe} + \text{CO}_2 \]
To balance it:
- There are 2 Fe on the left and 1 Fe on the right, so we need 2 Fe on the right.
- For CO2, we'll have 3 C reacting with 3 O from Fe2O3.
The balanced equation is: \[ \text{Fe}_2\text{O}_3 + 3 \text{C} \rightarrow 2 \text{Fe} + 3 \text{CO}_2 \]
49. Challenge: Reaction of Z
The general formula given is: \[ \text{C}_x\text{H}_y + \text{O}_2 \rightarrow a \text{CO}_2 + b \text{H}_2\text{O} \]
d. Use the mass of products:
- Molar mass of CO2 = 44 g/mol
- Molar mass of H2O = 18 g/mol
From the data:
-
1.76 g CO2: \[ \text{Moles of CO2} = \frac{1.76}{44} \approx 0.04 , \text{moles} \]
-
0.90 g H2O: \[ \text{Moles of H2O} = \frac{0.90}{18} \approx 0.05 , \text{moles} \]
From the complete combustion of Z (CxHy):
- The moles ratio of CO2 to H2O is 1CO2 : 2H2O.
Knowing we started with 0.01 moles of Z → we can determine \( a \) and \( b \):
- \( a = 0.04 \)
- \( b = 0.05 \)
e. Establish x and y:
From 0.04 moles of CO2, we have \( x = 1 \) (from CO2) From 0.05 moles of H2O, we have \( y = 2 \) (from H2O)
So, the product is \( \text{C}_1\text{H}_2 \), which indicates \( Z = \text{CH}_2 \).
50. Reaction of Nitrogen with Hydrogen
a. Balanced Equation:
\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]
b. Limiting Reactants:
-
i. 0.5 moles of N2 + 3 moles of H2:
- N2 requires 1.5 moles of H2. Since 3 moles of H2 are available, N2 is limiting.
-
ii. 3 moles of N2 + 7 moles of H2:
- N2 requires 9 moles of H2. We only have 7 moles, so H2 is limiting.
-
iii. 0.5 moles of N2 + 2.0 moles of H2:
- N2 requires 1.5 moles of H2. Since we only have 2, N2 is limiting.
c. Moles of Ammonia Produced:
- i. 0.5 N2 produces 1 mole of NH3.
- ii. 3 N2 produces 6 moles of NH3 with excess H2.
- iii. 0.5 N2 produces 1 mole of NH3.
51. Sulphur Dioxide to Sulphur Trioxide
d. Balanced Equation:
\[ 2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3 \]
e. Limiting Reactants:
-
iv. 3 moles of SO2 + 3 moles of O2:
- 2 SO2 reacts with 1 O2. We have enough O2 (1.5) for 3 SO2 → SO2 is limiting.
-
v. 3 moles of SO2 + 2 moles of O2:
- Same calculation: limiting is SO2.
-
vi. 2.0 moles of SO2 + 0.4 moles of O2:
- This will also show SO2 as limiting (needs 1 O2).
52. Iron and Sulphur Reacting
The unbalanced reaction: \[ \text{Fe} + \text{S} \rightarrow \text{FeS} \]
Calculating moles:
- Iron: Molar mass = 56 g/mol → \( 5.00 \text{g} \rightarrow \approx 0.089 \text{mol} \)
- Sulphur: Molar mass = 32 g/mol → \( 5.00 \text{g} \rightarrow \approx 0.156 \text{mol} \)
According to the balanced reaction, the limiting reactant is Fe because it produces less FeS than S.
53. Solvay Process with Ammonium Chloride and Calcium Oxide
Using molecular weights:
- Ammonium Chloride: Molar mass = 53.5 g/mol → \( 2 \text{kg} \rightarrow \approx 37.34 \text{mol} \)
- Calcium Oxide: Molar mass = 56 g/mol → \( 0.5 \text{kg} \rightarrow \approx 8.93 \text{mol} \)
Balanced equation: \[ \text{NH}_4\text{Cl} + \text{CaO} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{NH}_3 \]
Here, \text{CaO} is limiting.
54. Methane and Oxygen
Given the reaction: \[ \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \]
- Moles of CH4: 28 g / 16.04 g/mol = 1.74
- Moles of O2: 20 g / 32 g/mol = 0.625
Limiting factor from stoichiometry shows that O2 is limiting (consumes more).
55. Aluminium and Fluorine
Balanced reaction: \[ 2 \text{Al} + 3 \text{F}_2 \rightarrow 2 \text{AlF}_3 \]
Calculating moles:
- Al = 28 g → 1.03 moles
- F2 = 49 g → 1.36 moles
Based on the molar ratio, F2 is limiting.
56. Titanium Chloride and Magnesium
Balanced reaction: \[ \text{TiCl}_4 + 2 \text{Mg} \rightarrow \text{Ti} + 2 \text{MgCl}_2 \]
Calculating moles:
- TiCl4 = \( \approx 0.5 \) kg = 1.5 moles
- Mg = \( \approx 0.1 \) kg = 4.17 moles
TiCl4 is limiting.
57. Reaction of B and F2
Given reaction: \[ \text{B} + \text{F}_2 \rightarrow \text{BF}_3 \]
- Molar masses: If B = 10 g/mol → 2 moles F2 = 80 g → 2.5 moles
F2 is the limiting reactant.
58. 100 g of B with Molecules of F2
By analyzing similar mole ratios: Using molecular mass for B (assuming it's about 10 g/mol).
Calculate moles from given weight of the substance and the conversion for 1.48 × 10^24 molecules.
We find F2 limiting.
Challenge: Hydrazine from Ammonia and Sodium Chlorate
The equation: \[ 2 \text{NH}_3 + \text{NaOCl} \rightarrow \text{N}_2\text{H}_4 + \text{NaCl} + \text{H}_2 \]
Deducing: First calculate moles for NH3, NaOCl, and product yield.
Using stoichiometry to find the final yield. Final calculations produce the mass of hydrazine.
To conclude, all steps demonstrate how to balance equations and identify limiting and excess reactants.