2Na + 2H2O → 2NaOH + H2
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Word Equation
Sodium + Water = Sodium Hydroxide + Dihydrogen
Two moles of
and two moles of
react đồ sộ form two moles of
and one mole of
For sodium khuông into the cup distilled water → Sodium reacts with water, melt the white circle drops fast moving on the water. Na melting sample for the rest, there flying H2 gas, reaction heat radiating. Water evaporating liquid khuông, will be a white solid, which is sodium Hidr
Reactants
Sodium - Na
Natrium Element 11
Water - H2O
Hydroxic Acid H₂O [Oh2] Aqua Pure Water Hydrogen Oxide Oxidane Dihydrogen Oxide
Products
Sodium Hydroxide - NaOH
White Caustic Naoh Sodium Hydrate Soda Lye Lye Caustic Soda
Dihydrogen - H2
Molecular Hydrogen Hydrogen Molecule Hydrogen Hydrogen Gas Molecular Hydrogen Gas H₂
Thermodynamics
Thermodynamics of the reaction can be calculated using a lookup table.
Is the Reaction Exothermic or Endothermic?
| Na (s) | 2 mol | 0 kJ/mol | -0 kJ |
|---|---|---|---|
| H2O (g) | 2 mol | -241.818464 kJ/mol | 483.636928 kJ |
| NaOH (l) | 2 mol | -416.89376 kJ/mol | -833.78752 kJ |
| H2 (g) | 1 mol | 0 kJ/mol | 0 kJ |
| ΣΔH°f(reactants) | -483.636928 kJ | ||
| ΣΔH°f(products) | -833.78752 kJ | ||
| ΔH°rxn | -350.150592 kJ | ||
ΣΔH°f(reactants) > ΣΔH°f(products), ví Na + H2O = NaOH + H2 is exothermic (releases heat).
Is the Reaction Exoentropic or Endoentropic?
ΔS = Sproducts - Sreactants. If ΔS < 0, it is exoentropic. If ΔS > 0, it is endoentropic.
| Na (s) | 2 mol | 51.4632 J/(mol K) | -102.9264 J/K |
|---|---|---|---|
| H2O (g) | 2 mol | 188.715136 J/(mol K) | -377.430272 J/K |
| NaOH (l) | 2 mol | 75.85592 J/(mol K) | 151.71184 J/K |
| H2 (g) | 1 mol | 130.586824 J/(mol K) | 130.586824 J/K |
| ΣΔS°(reactants) | 480.356672 J/K | ||
| ΣΔS°(products) | 282.298664 J/K | ||
| ΔS°rxn | -198.058008 J/K | ||
ΣΔS°(reactants) > ΣΔS°(products), ví Na + H2O = NaOH + H2 is exoentropic (decrease in entropy).
Is the Reaction Exergonic or Endergonic?
ΔG = Gproducts - Greactants. If ΔG < 0, it is exergonic. If ΔG > 0, it is endergonic.
| Na (s) | 2 mol | 0 kJ/mol | -0 kJ |
|---|---|---|---|
| H2O (g) | 2 mol | -228.588656 kJ/mol | 457.177312 kJ |
| NaOH (l) | 2 mol | -374.13328 kJ/mol | -748.26656 kJ |
| H2 (g) | 1 mol | 0 kJ/mol | 0 kJ |
| ΣΔG°(reactants) | -457.177312 kJ | ||
| ΣΔG°(products) | -748.26656 kJ | ||
| ΔG°rxn | -291.089248 kJ | ||
ΣΔG°(reactants) > ΣΔG°(products), ví Na + H2O = NaOH + H2 is exergonic (releases energy).
Reaction Expressions
Kc or Q = ( [NaOH]2 [H2] ) / ( [Na]2 [H2O]2 )
(assuming all reactants and products are aqueous. substitutue 1 for any solids/liquids, and Psubstance for gases.)
rate = -1/2 * (Δ[Na] / Δt) = -1/2 * (Δ[H2O] / Δt) = một nửa * (Δ[NaOH] / Δt) = (Δ[H2] / Δt)
(assuming constant volume in a closed system and no accumulation of intermediates or side products)
Instructions
To balance a chemical equation, enter an equation of a chemical reaction and press the Balance button. The balanced equation will appear above.
Xem thêm: fe203 + hcl
- Use uppercase for the first character in the element and lowercase for the second character. Examples: Fe, Au, Co, Br, C, O, N, F.
- Ionic charges are not yet supported and will be ignored.
- Replace immutable groups in compounds đồ sộ avoid ambiguity. For example, C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but XC2H5 + O2 = XOH + CO2 + H2O will.
- Compound states [like (s) (aq) or (g)] are not required.
- You can use parenthesis () or brackets [].
How To Balance Equations
Read our article on how đồ sộ balance chemical equations or ask for help in our chat.
You can also ask for help in our chat or forums.
Practice Balancing Na + H2O = NaOH + H2
Balance Na + H2O = NaOH + H2 Using the Algebraic Method
To balance the equation Na + H2O = NaOH + H2 using the algebraic method step-by-step, you must have experience solving systems of linear equations. The most common methods are substitution/elimination and linear algebra, but any similar method will work.
Step 1: Label Each Compound With a Variable
Label each compound (reactant or product) in the equation with a variable đồ sộ represent the unknown coefficients.
a Na + b H2O = c NaOH + d H2
Step 2: Create a System of Equations
Create an equation for each element (Na, H, O) where each term represents the number of atoms of the element in each reactant or product.
Na: 1a + 0b = 1c + 0d H: 0a + 2b = 1c + 2d O: 0a + 1b = 1c + 0d
Step 3: Solve For All Variables
Use substitution, Gaussian elimination, or a calculator đồ sộ solve for each variable.
- 1a - 1c = 0
- 2b - 1c - 2d = 0
- 1b - 1c = 0
Use your graphing calculator's rref() function (or an online rref calculator) đồ sộ convert the following matrix into reduced row-echelon-form:
[ 1 0 -1 0 0] [ 0 2 -1 -2 0] [ 0 1 -1 0 0]
The resulting matrix can be used đồ sộ determine the coefficients. In the case of a single solution, the last column of the matrix will contain the coefficients.
Simplify the result đồ sộ get the lowest, whole integer values.
- a = 2 (Na)
- b = 2 (H2O)
- c = 2 (NaOH)
- d = 1 (H2)
Step 4: Substitute Coefficients and Verify Result
Count the number of atoms of each element on each side of the equation and verify that all elements and electrons (if there are charges/ions) are balanced.
2 Na + 2 H2O = 2 NaOH + H2
| Na | 2 | 2 | ✔️ |
|---|---|---|---|
| H | 4 | 4 | ✔️ |
| O | 2 | 2 | ✔️ |
Since there is an equal number of each element in the reactants and products of 2Na + 2H2O = 2NaOH + H2, the equation is balanced.
Balance Na + H2O = NaOH + H2 Using Inspection
The law of conservation of mass states that matter cannot be created or destroyed, which means there must be the same number atoms at the over of a chemical reaction as at the beginning. To be balanced, every element in Na + H2O = NaOH + H2 must have the same number of atoms on each side of the equation. When using the inspection method (also known as the trial-and-error method), this principle is used đồ sộ balance one element at a time until both sides are equal and the chemical equation is balanced.
Step 1: Count the number of each element on the left and right hand sides
| Reactants (Left Hand Side) | Products (Right Hand Side) | ||||||
|---|---|---|---|---|---|---|---|
| Reactants | Products | ||||||
| Na | H2O | Total | NaOH | H2 | Total | ||
| Na | 1 | 1 | 1 | 1 | ✔️ | ||
| H | 2 | 2 | 1 | 2 | 3 | ❌ | |
| O | 1 | 1 | 1 | 1 | ✔️ | ||
Step 2: Multiply coefficients for compounds đồ sộ balance out each element
For each element that is not equal, try đồ sộ balance it by adding more of it đồ sộ the side with less. Sometimes there may be multiple compounds with that element on one side, ví you'll need đồ sộ use your best judgement and be prepared đồ sộ go back and try the other options.
H is not balanced. Add 1 molecule of H2O đồ sộ the reactant (left-hand) side đồ sộ try đồ sộ balance Hydrogen:
Na + 2H2O = NaOH + H2Reactants Products Na 1 1 ✔️ H 4 3 ❌ O 2 1 ❌ H is not balanced. Add 1 molecule of NaOH đồ sộ the product (right-hand) side đồ sộ balance Hydrogen:
Na + 2H2O = 2NaOH + H2Reactants Products Na 1 2 ❌ H 4 4 ✔️ O 2 2 ✔️ Na is not balanced. Add 1 molecule of Na đồ sộ the reactant (left-hand) side đồ sộ balance Sodium:
2Na + 2H2O = 2NaOH + H2Reactants Products Na 2 2 ✔️ H 4 4 ✔️ O 2 2 ✔️
Step 3: Verify that the equation is balanced
Since there are an equal number of atoms of each element on both sides, the equation is balanced.
2Na + 2H2O = 2NaOH + H2
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