Answer:
2 moles
Explanation:
The following were obtained from the question:
Molarity = 0.25 M
Volume = 8L
Mole =?
Molarity is simply defined as the mole of solute per unit litre of solution. It is represented mathematically as:
Molarity = mole of solute/Volume of solution.
With the above equation, we can easily find the number of mole of MgCl2 present in 8 L of 0.25 M MgCl2 solution as follow:
Molarity = mole of solute/Volume of solution.
0.25 = mole of MgCl2 /8
Cross multiply to express in linear form
Mole of MgCl2 = 0.25 x 8
Mole of MgCl2 = 2 moles
Therefore, 2 moles of MgCl2 are present in 8 L of 0.25 M MgCl2 solution
You can calculate the excess reactant by subtracting the mass of excess reagent consumed from the total mass of reagent given therefore,
The answer: Theoretical yield is 121.60 g of NH₃
Excess reactant is H₂
Rate limiting reactant is N₂
explanation: 100 g of Nitrogen
100 g of hydrogen
We are required to identify the theoretical yield of the reaction, the excess reactant and the rate limiting reagent.
We first write the equation for the reaction between nitrogen and hydrogen;
N₂ + 3H₂ → 2NH₃
From the reaction 1 mole of nitrogen reacts with 3 moles of Hydrogen gas.
Secondly we determine the moles of nitrogen gas given and hydrogen gas given;
Moles of Nitrogen gas
Moles = Mass ÷ Molar mass
Molar mass of nitrogen gas = 28.0 g/mol
Moles of Nitrogen gas = 100 g ÷ 28 g/mol 3.57 moles
Moles of Hydrogen gas
Molar mass of Hydrogen gas = 2.02 g/mol
Moles = 100 g ÷ 2.02 g/mol
= 49.50 moles
From the mole ratio given by the equation, 1 mole of nitrogen requires 3 moles of Hydrogen gas.
Thus, 3.57 moles of Nitrogen gas requires (3.57 × 3) 10.71 moles of Hydrogen gas.
This means, Nitrogen gas is the rate limiting reagent and hydrogen gas is the excess reactant.
Third calculate the theoretical yield of the reaction.
1 mole of nitrogen reacts to from 2 moles of ammonia gas
Therefore;
Moles of ammonia gas produced = Moles of nitrogen × 2
= 3.57 moles × 2
= 7.14 moles
But; molar mass of Ammonia gas is = 17.03 g/mol
Therefore;
Mass of ammonia gas produced = 7.14 moles × 17.03 g/mol
= 121.59 g
= 121.60 g
Thus, the theoretical amount of ammonia gas produced is 121.60 g
The major species in solution when solid ammonium bromate is dissolved in water is shown below
Answer:
2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.
Explanation:
- For balancing a chemical equation, we should apply the law of conversation of mass. It states that the no. of atoms in the reactants side is equal to that of the products side.
So, the balanced equation:
<em>2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.</em>
It is clear that 2.0 moles of C₃H₇BO₃ is completely burned in 8 m oles of oxygen and produce 6 moles of CO₂, 7 moles of H₂O and 1 mole of B₂O₃.
The balanced equation for the above neutralisation reaction is as follows;
Ca(OH)₂ + 2HCl ----> CaCl₂ + 2H₂O
Stoichiometry of Ca(OH)₂ to HCl is 1:2
number of Ca(OH)₂ moles reacted - 0.250 mol/L x 20.0 x 10⁻³ L = 5.00 x 10⁻³ mol
according to molar ratio of 1:2
number of HCl moles required = 2 x number of Ca(OH)₂ moles reacted
number of HCl moles = 5.00 x 10⁻³ x 2 = 10.0 x 10⁻³ mol
molarity of HCl solution - 0.250 M
there are 0.250 mol in volume of 1 L
therefore 10.0 x 10⁻³ mol in - 10.0 x 10⁻³ mol / 0.250 mol/L = 40.0 mL
40.0 mL of 0.250 M HCl is required
