Cao + H2O ---->Ca(OH)2
Calculate the number of each reactant and the moles of the product
that is
moles = mass/molar mass
The moles of CaO= 56.08g/ 56.08g/mol(molar mass of Cao)= 1mole
the moles of water= 36.04 g/18 g/mol= 2.002moles
The moles of Ca (OH)2=74.10g/74.093g/mol= 1mole
The mass of differences of reactant and product can be therefore
explained as
1 mole of Cao reacted completely with 1 mole H2O to produce 1 mole of Ca(OH)2. The mass of water was in excess while that of CaO was limited
Given the model from the question,
- The products are: N₂, H₂O and H₂
- The reactants are: H₂ and NO
- The limiting reactant is H₂
- The balanced equation is: 3H₂ + 2NO —> N₂ + 2H₂O + H₂
<h3>Balanced equation </h3>
From the model given, we obtained the ffolowing
- Red => Oxygen
- Blue => Nitrogen
- White => Hydrogen
Thus, we can write the balanced equation as follow:
3H₂ + 2NO —> N₂ + 2H₂O + H₂
From the balanced equation above,
- Reactants: H₂ and NO
- Product: N₂, H₂O and H₂
<h3>How to determine the limiting reactant</h3>
3H₂ + 2NO —> N₂ + 2H₂O + H₂
From the balanced equation above,
3 moles of H₂ reacted with 2 moles of NO.
Therefore,
5 moles of H₂ will react with = (5 × 2) / 3 = 3.33 moles of NO
From the calculation made above, we can see that only 3.33 moles of NO out of 4 moles given are required to react completely with 5 moles of H₂.
Thus, H₂ is the limiting reactant
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As we know,
1 D = 3.34 × 10⁻³⁰ C.m
So,
1.44 D = ?
Solving for 1.44 D,
= (3.34 × 10⁻³⁰ C.m × 1.44 D) ÷ 1 D
1.44 D = 4.80 × 10⁻³⁰ C.m
Dipole Moment is given as,
Dipole Moment = q × r
Solving for q,
q = Dipole Moment / r ------ (1)
Where,
Dipole Moment = 4.80 × 10⁻³⁰ C.m
r = 163 pm = 1.63 × 10⁻¹⁰ m
Putting values in eq. 1,
q = 4.80 × 10⁻³⁰ C.m / 1.63 × 10⁻¹⁰ m
q = 2.94 × 10⁻²⁰ C
As,
1.602 × 10⁻¹⁹ C = 1 e⁻
So,
2.94 × 10⁻²⁰ C = X e⁻
Solving for X,
X = (2.94 × 10⁻²⁰ C × 1 e⁻) ÷ 1.602 × 10⁻¹⁹ C
= 0.183 e⁻
Result:
So one element is containing + 0.183 e⁻ while the other element is containing - 0.183 e⁻.
Help 2 what the answer for this?
(A)Nuclear change..............
