7.5 mol of hydrogen would be needed to consume the available nitrogen.
Explanation:
When hydrogen reacts with nitrogen, ammonia is formed as shown below;
3H₂ (g) + N₂ (g) → 2NH₃ (g)
As seen from the equation, every 3 moles of H₂ react with a mole of N₂ to form 2 moles of NH₃.
The limiting factor in a chemical reaction is the reactant that gets depleted first.
Because the molar mass of nitrogen gas is approximately 28g/mol, 70g of nitrogen gas would be 2.5 moles.
The reaction ratio of nitrogen to hydrogen in the reaction is 1 : 3. The reaction would require 2.5 * 3 (7.5) moles of hydrogen for a complete reaction.
However since there are only 7g on hydrogen, (Remember 1 mole of H₂ is approximately 2g), the available moles of H₂ is 7 / 2 = 3.5
3.5 moles fall short of the 7.5 moles of H₂ required for a complete reaction. H₂ gets depleted first before N₂. The reaction would require 4 more moles of H₂.
I mole of water has an Avogadro number of molecules.
1 mole = 6.02 * 10^ 23 molecules.
6.02 * 10^ 23 molecules = 1 mole of water
1 molecule = 1/(6.02 * 10^23) mole of water
2.0 * 10^22 molecules would have = (2*10^22) * 1/(6.02*10^23)
= 0.033
2* 10 ^22 molecules of water would have 0.033 moles of water.
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X
Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²
Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M
Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L
Molar mass of CaCO₃ = 100 g mol⁻¹
Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L
Your correct answer is C. 97%
