Answer:
Nitrogen is limiting reactant while hydrogen is in excess.
Explanation:
Given data:
Mass of N₂ = 25 g
Mass of H₂ = 25 g
Mass of ammonia formed = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of Nitrogen:
Number of moles = mass/ molar mass
Number of moles = 25 g/ 28 g/mol
Number of moles = 0.89 mol
Number of moles of hydrogen:
Number of moles = mass/ molar mass
Number of moles = 25 g/ 2 g/mol
Number of moles = 12.5 mol
Now we will compare the moles of both reactant with ammonia.
H₂ ; NH₃
3 : 2
12.5 : 2/3×12.5 = 8.3
N₂ ; NH₃
1 : 2
0.89 : 2×0.89 = 1.78
The number of moles of ammonia produced by nitrogen are less thus nitrogen is limiting reactant while hydrogen is in excess.
37.8 g CH2Br2 X (1 mol CH2Br2 / 173.83 g) = 4.60X10^-3 mol CH2Br2
<span>4.60X10^-3 mol CH2Br2 X (2 mol Br / 1 mol CH2Br2) X 6.02X10^23 atoms/mol = 5.54X10^21 bromine atoms.
I think this is the answer.</span>
Both active and passive transport move molecules
So if we use the equation:
→ 
We can then determine the amount of 
needed to produce 208 kg of methanol.
So let's find out how many moles of methanol 208 kg is:
Methanol molar weight = 32.041g/mol
So then we can solve for moles of methanol:
So now that we have the amount of moles produced, we can use the molar ratio (from the balanced equation) of hydrogen and methanol. This ratio is 2:1 hydrogen:methanol.
Therefore, we can set up a proportion to solve for the moles of hydrogen needed:
So now that we have the number of moles of 
that are produced, we can then use the molar weight of hydrogen to solve for the mass that is needed:
Therefore, the amount of diatomic hydrogen () that is needed to produce 208kg of methanol is 
g.
Answer:
Dirty sand is piled on a sheet of fine mesh stretched between two long poles, the mesh collects the mircoplastic and other materials while allowing the sand to fall through.
