Answer:
0.56 g
Explanation:
<em>A chemist determines by measurements that 0.020 moles of nitrogen gas participate in a chemical reaction. Calculate the mass of nitrogen gas that participates.</em>
Step 1: Given data
Moles of nitrogen gas (n): 0.020 mol
Step 2: Calculate the molar mass (M) of nitrogen gas
Molecular nitrogen is a gas formed by diatomic molecules, whose chemical formula is N₂. Its molar mass is:
M(N₂) = 2 × M(N) = 2 × 14.01 g/mol = 28.02 g/mol
Step 3: Calculate the mass (m) corresponding to 0 0.020 moles of nitrogen gas
We will use the following expression.
m = n × M
m = 0.020 mol × 28.02 g/mol
m = 0.56 g
Answer : The correct option is (3) 500 K and 0.1 atm.
Explanation :
A real gas behaves ideally at high temperature and low pressure.
The ideal gas equation is,
where,
P = pressure of gas
V = Volume of gas
R = Gas constant
T = temperature of gas
n = number of moles of gas
The ideal gas works properly when the inter-molecular interactions between the gas molecules and volume of gas molecule will be negligible. This is possible when pressure is low and temperature is high.
Therefore, the correct option is (3) 500 K and 0.1 atm.
Answer:
Mass = 357.7 g
Explanation:
Given data:
Mass of Fe = 250 g
Mass of oxygen = 120 g
Mass of iron(III) oxide produced = ?
Solution:
Chemical equation:
4Fe + 3O₂ → 2Fe₂O₃
Number of moles of Fe:
Number of moles = mass/molar mass
Number of moles = 250 g/ 55.8 g/mol
Number of moles = 4.48 mol
Number of moles of O₂ :
Number of moles = mass/molar mass
Number of moles = 120 g/ 32 g/mol
Number of moles = 3.75 mol
Now we will compare the moles of reactants with product.
Fe : Fe₂O₃
4 : 2
4.48 : 2/4×4.48 = 2.24
O₂ : Fe₂O₃
3 : 2
3.75 : 2/3×3.75= 2.5
Less number of moles of Fe₂O₃ are produced by Fe thus it will act as limiting reactant.
Mass of Fe₂O₃:
Mass = number of moles × molar mass
Mass = 2.24 mol × 159.69 g/mol
Mass = 357.7 g
Answer:
5 moles of oxygen are required.
Explanation:
Given data:
Moles of O₂ required = ?
Moles of H₂ present = 10 mol
Solution:
Chemical equation:
O₂ + 2H₂ → 2H₂O
Now we will compare the moles of oxygen and hydrogen.
H₂ : O₂
2 : 1
10 : 1/2×10 = 5 mol
5 moles of oxygen are required.
