Answer:
R.F.M of Iron (II) oxide :
Moles :
Molecules :
Answer:
30.4 g. NH3
Explanation:
This problem tells us that the hydrogen (H2) is the limiting reactant, as there is "an excess of nitrogen." Using stoichiometry (the relationship between the various species of the equation), we can see that for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
But before we can use that relationship to find the number of grams of ammonia produced, we need to convert the given grams of hydrogen into moles:
5.4 g x [1 mol H2/(1.008x2 g.)] = 2.67857 mol H2 (not using significant figures yet; want to be as accurate as possible)
Now, we can use the relationship between H2 and NH3.
2.67857 mol H2 x (2 mol NH3/3 mol H2) = 1.7857 mol NH3
Now, we have the number of moles of ammonia produced, but the answer asks us for grams. Use the molar mass of ammonia to convert.
1.7857 mol NH3 x 17.034 g. NH3/mol NH3 = 30.4 g. NH3 (used a default # of 3 sig figs)
1. Molarity : 0.25 M
2. mol CH₄ = 7.4 moles
mol O₂ = 14.8 moles
<h3>Further explanation</h3>
1.
Given
83.2 g CuCl2 in 2.5 liters of water
Required
the molarity
Solution
Molarity : mol solute per liter of solution(not per liter of solvent)
mol solute = mol CuCl₂
mol CuCl₂ = mass : MW CuCl₂
mol CuCl₂ = 83.2 : 134.45
mol CuCl₂ = 0.619
Molarity(M) = mol : V
Assume density CuCl₂ = 3.39 g/cm³
volume CuCl₂ = 8.32 g : 3.39 g/cm³ = 2.45 cm³=2.45 x 10⁻³ L
With this small volume value of CuCl₂, the volume of the solute is sometimes neglected in calculating molarity
volume of solution = 2.5 L + 2.45 x 10⁻³ L = 2.50245 L
Molarity(M) = mol : V
M = 0.619 : 2.50245 L = 0.247≈0.25
2.
Given
Reaction
The correct balanced reaction:
CH4 + 2O2 → CO2 + 2H2O
7.4 moles CO2
Required
moles of methane (CH4) and oxygen gas (O2)
Solution
From the equation, mol ratio of CO₂ : CH₄ : O₂ = 1 : 1 : 2
mol CH₄ = mol CO₂ = 7.4 moles
mol O₂ = 2 x mol CO₂ = 2 x 7.4 moles = 14.8 moles
The net ionic equation is as below
H^+(aq) + H^-(aq) → H2O(l)
Explanation
from the chemical equation
HClO4(aq) + NaOH (aq) → H2O(l) + NaClO4(aq)
write ionic equation
that is,
H^+(aq)+ClO4^-(aq) + Na^+(aq) +OH^-(aq)→ H2O(l) + Na^+(aq)+ClO4^-(aq)
cancel the spectator ions in both side(ions that do no take place in a chemical equation)
that ClO4^- and Na^+
the net ionic is therefore
= H^+(aq) + OH^-(aq) → H2O(l)