Answer:
To answer the question, we correctly fill the attached screenshot as follows;
- 3H₂ + N₂ → 2NH₃
- The molar mass of H₂ = 2 g/mol
The molar mass of N₂ = 28 g/mol
A. Therefore, the excess reactant is hydrogen gas H₂ because it makes the most amount of ammonia, NH₃ (33. moles of NH₃)
B. The limiting reactant in nitrogen, N₂, because it is the reactant that makes the least amount of the ammonia, NH₃ (approximately 7.143 mol NH₃)
C. The theoretical yield of ammonia, is the maximum amount of ammonium that can be produced from the reaction between the 100 g of hydrogen gas, H₂, and 100 g of nitrogen gas, N₂ which is given by the amount of ammonia produced by the limiting reactant which is approximately 7.143 mole of NH₃
Explanation:
Answer:
1-chlorobutane
Explanation:
Given that :
-- The compound A is made to react with alcoholic KOH and produce compound B.
-- The compound B on Ozonolysis produces methanol as well as propanol
The compound A should be a haloalkanes as treatment of haloalkanes with the alcoholic KOH, it will give alkene.
So, the compound B should be Butene
Now the ozonolysis of the compound B or butene will give methanol and propanal as shown :
Therefore, the compound A is 1-chlorobutane.
SrCl₂ reacts with conc. H₂SO₄ and gives SrSO₄ and 2HCl as the products. The balanced equation is
SrCl₂<span> + H</span>₂SO₄<span> → SrSO</span>₄<span> + 2HCl
moles = mass / molar mass
molar mass of </span>H₂SO₄ = 98 g mol⁻¹
moles of H₂SO₄ = 300.0 g / 98 g mol⁻¹ = 3.06 mol
Stoichiometric ratio between SrCl₂ and H₂SO₄ is 1 : 1
Hence, moles of SrCl₂ = moles of H₂SO₄
= 3.06 mol
molar mass of SrCl₂ = 158.53 g mol⁻¹
Hence, mass of SrCl₂ = 3.06 mol x 158.53 g mol⁻¹
= 485.1 g
Hence, the mass of SrCl₂ needed to react with H₂SO₄ is 485.1 g
Honestly just look at the question deeper and find it inside the equation of the fraction