<span>Answer:
Equation:
N2(g) + 3H2(g) → 2NH3(g)
Check for limiting reactant
Molar mass N2 = 28g/mol
mol N2 in 31,500g = 31,500/28 = 1,125 mol N2
This will require 1,125*3 = 3,375 mol H2
Molar mass H2 = 2g/mol
Mol H2 in 5,220g = 5,220/2 = 2,610 mol H2
The H2 is limiting
From the equation:
3 mol H2 will produce 2mol NH3
2610 mol H2 will produce 2610*2/3 = 1,740 mol NH3
Molar mass NH3 = 17g/mol
Mass of 1740 mol NH3 = 1740*17 = 29,580g NH3 produced
Mass of NH3 produced = 29.6kg</span>
Iron(III) oxide = Fe₂O₃
Carbon = C
Carbon monoxide = CO
Iron = Fe
Fe₂O₃ + C ⇒ CO + Fe
These are the correct formulas, but the equation is not balanced. There are two Fe's on the left side, so we have to put a 2 in front of the Fe on the right. Also, there are 3 O's on the left but only 1 on the right, so we need to put a 3 in front of the CO; however, this 3 makes it 3 C's on the right, so we balance that off by putting a 3 in front of the C on the left:
Fe₂O₃ + 3C ⇒ 3CO + 2Fe
The empirical formula can be C₄H₈ or CH₃-CH=CH-CH₃.
Explanation:
As the number of moles of each element is already given ,then there is no need of further calculations. We can get the empirical formula by writing the moles of each element in its subscript position. As empirical formula is one of the simplest way to represent any chemical compound. So the given compound can be written as CₓHₙ. Here x and n are the number of moles of carbon and hydrogen present in the compound.
Since it is stated as the 4 moles of Carbon and 8 moles of hydrogen is present and it is known that each carbon can have maximum four single bond with its neighbors, so the empirical formula can be written as
C₄H₈ or CH₃-CH=CH-CH₃
Thus, the empirical formula can be C₄H₈ or CH₃-CH=CH-CH₃.
Answer:
c is speed of light (in terms of velocity)
v is velocity
lambda is wavelength
Explanation:
speed of light is constant at 3 x 10^8 m/s which is a velocity value. Lambda is usually in the units of nanometers which represents wavelength.
Answer:
Explanation:
C₆H₁₂O₆ + 60₂ = 6H₂0 + 6CO₂.
1 mole 6 mole
molecular weight of C₆H₁₂O₆ = 180 g
molecular weight of oxygen = 32 g
1 gram of glucose = 1 / 180 = 5.55 x 10⁻³ moles
1 gram of oxygen = 1 / 32 = 31.25 x 10⁻³ moles
1 mole of glucose reacts with 6 moles of oxygen
5.55 x 10⁻³ moles of glucose will react with 6 x 5.55 x 10⁻³ moles of oxygen
= 33.30 x 10⁻³ moles of oxygen .
But oxygen available = 31.25 x 10⁻³ moles
So available oxygen is less than required .
Hence oxygen is the limiting reagent .
b ) is the right option .