find mol of N2 present using gas law equation
PV = nRT
P = pressure = 688/760 = 0.905 atm.
V = 100mL = 0.1L
n = ???
R = 0.082057
T = 565+273 = 838
Substitute:
0.905*0.1 = n*0.082057*838
n = 0.0905 / 68.76
n = 0.00132 mol N2
Molar mass N2 = 28 g/mol
0.00132 mol = 0.00132*28 = 0.037g N2 gas
One mole of Fe(NO3)3, or iron(III) nitrate, has three moles of nitrate molecules, which have three moles of oxygen atoms each. We can show this mathematically:
1 mole Fe(NO3)3 * (3 moles NO3)/(1 mole Fe(NO3)3) = 3 moles NO3
3 moles NO3 * (3 moles Oxygen)/(1 mole NO3) = 9 moles Oxygen
9 moles of Oxygen in one mole Fe(NO3)3
Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Answer:
Theoretical yield of C6H10 = 3.2 g.
Explanation:
Defining Theoretical yield as the quantity of product obtained from the complete conversion of the limiting reactant in a chemical reaction. It can be expressed as grams or moles.
Equation of the reaction
C6H11OH --> C6H10 + H2O
Moles of C6H11OH:
Molar mass of C6H110H = (12*6) + (1*12) + 16
= 100 g/mol
Mass of C6H10 = 3.8 g
number of moles = mass/molar mass
=3.8/100
= 0.038 mol.
Using stoichoimetry, 1 moles of C6H110H was dehydrated to form 1 mole of C6H10 and 1 mole of water.
Therefore, 0.038 moles of C6H10 was produced.
Mass of C6H10 = molar mass * number of moles
Molar mass of C6H10 = (12*6) + (1*10)
= 82 g/mol.
Mass = 82 * 0.038
= 3.116 g of C6H10.
Theoretical yield of C6H10 = 3.2 g
