Answer:
The number ratio is 4:7
Explanation:
Step 1: Data given
Compound 1 has 50.48 % oxygen
Compound 2 has 36.81 % oxygen
Molar mass oxygen = 16 g/mol
Molar mass manganese = 54.94 g/mol
Step 2: Calculate % manganes
Compound 1: 100 - 50.48 = 49.52 %
Compound 2: 100 - 36.81 = 63.19 %
Step 3: Calculate mass
Suppose mass of compounds = 100 grams
Compound 1:
50.48 % O = 50.48 grams
49.52 % Mn = 49.52 grams
Compound 2:
36.81 % O = 36.81 grams
63.19 % Mn = 63.19 grams
Step 4: Calculate moles
Compound 1
Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles
Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles
Compound 2
Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles
Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles
Step 5: calculate mol ratio
We will divide by the smallest amount of moles
Compound 1
O: 3.155/0.9013 = 3.5
Mn: 0.9013 / 0.9013 = 1
Mn2O7
Compound 2
O: 2.301 / 1.150 = 2
Mn: 1.150 / 1.150 = 1
MnO2
The number ratio is 2:3.5 or 4:7
Answer:
2 moles
Explanation:
In one mole of O2 there are 16 grams. So in 2 moles there are 32 grams
Answer:
oxygen is limiting reactant
Explanation:
Given data:
Mass of phosphorus = 25.0 g
Mass of oxygen = 50.0 g
What is limiting reactant ?
Solution:
Chemical equation:
P₄ + 5O₂ → P₄O₁₀
Number of moles of P₄:
Number of moles = mass/molar mass
Number of moles = 25.0 g/ 123.89 g/mol
Number of moles = 0.20 mol
Number of moles of O₂:
Number of moles = mass/molar mass
Number of moles = 50.0 g/ 32 g/mol
Number of moles = 1.56 mol
now we will compare the moles of reactants with product:
P₄ : P₄O₁₀
1 : 1
0.20 : 0.20
O₂ : P₄O₁₀
5 : 1
1.56 : 1/5×1.56 = 0.312 mol
Less number of moles of product are formed by the oxygen thus it will act as limiting reactant.
