NH2CH2COOH + Na2CO3 = NH2CH2COONa + H2O + CO2 ...

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Balanced Chemical Equation

2NH2CH2COOH + Na2CO3 → 2NH2CH2COONa + H2O + CO2 Reaction Info Balance Another Equation Step-by-Step Solution Practice Balancing

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Word Equation

Glycine + Sodium Carbonate = Sodium Glycinate + Water + Carbon Dioxide

Two moles of Glycine [NH2CH2COOH] and one mole of Sodium Carbonate [Na2CO3] react to form two moles of Sodium Glycinate [NH2CH2COONa], one mole of Water [H2O] and one mole of Carbon Dioxide [CO2]

Reactants

• Glycine - NH2CH2COOH

  Hgly H2N-Ch2-Cooh Aminoacetic Acid Glycocoll Glykokoll Aminoethanoic Acid Glycin Glyzin NH2CH2COOH Molar Mass NH2CH2COOH Oxidation Number

• Sodium Carbonate - Na2CO3

  Soda Sodium Carbonate (Anhydrous) Sodium Salt Of Carbonic Acid Sodium Carbonate Anhydrous Natrum Carbonicum Soda-Ash Disodium Carbonate Sodium Carbonate, Anhydrous Na2CO3 Molar Mass Na2CO3 Oxidation Number

Products

• Sodium Glycinate - NH2CH2COONa

• Water - H2O

  Hydroxic Acid H₂O [Oh2] Aqua Pure Water Hydrogen Oxide Oxidane Dihydrogen Oxide H2O Molar Mass H2O Oxidation Number

• Carbon Dioxide - CO2

  Co₂ Carbon Oxide Methanedione Carbon-12 Dioxide Carbonic Acid Anhydride E290 E 290 E-290 CO2 Molar Mass CO2 Oxidation Number

Show Chemical Structure Image

Reaction Expressions

Equilibrium Constant & Reaction Quotient Kc or Q = ( [NH2CH2COONa]2 [H2O] [CO2] ) / ( [NH2CH2COOH]2 [Na2CO3] )(assuming all reactants and products are aqueous. substitutue 1 for any solids/liquids, and Psubstance for gases.) Rate of Reaction rate = -1/2 * (Δ[NH2CH2COOH] / Δt) = -(Δ[Na2CO3] / Δt) = 1/2 * (Δ[NH2CH2COONa] / Δt) = (Δ[H2O] / Δt) = (Δ[CO2] / Δt)(assuming constant volume in a closed system and no accumulation of intermediates or side products) Calculate Reaction Stoichiometry Calculate Limiting Reagent Chemical Equation Balancer 🛠️ Balance Equation ➜

Instructions

To balance a chemical equation, enter an equation of a chemical reaction and press the Balance button. The balanced equation will appear above.

• 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 to 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

Balance any equation or reaction using this chemical equation balancer! Find out what type of reaction occured.

Read our article on how to balance chemical equations or ask for help in our chat.

Examples

• Glycine + Sodium Carbonate = Sodium Glycinate + Water + Carbon Dioxide
• NH2CH2COOH + Na2CO3 = NH2CH2COONa + CO2 + H2O
• NH2CH2COOH + Na2CO3 = NH2CH2COONa + H2CO3
• K2Cr2O7 + H2O2 = Cr2O3 + KOH + O2 + H2O
• NaClO3 + P2S5 + H2O = H3PO4 + H2SO4 + NaCl
• CO2 + CsOH = CCs2O3 + H2O
• LiNO3 + NaOH = NaNO3 + HLiO
• Cl2 + SO3 = ClSO3
• FeS + BaCl = FeCl + BaS
• NaNO3 + Na2SO4 = Na2SO4 + NaNO3
• FeCr2O4 + O2 + Na2CO3 + CaO = Fe2O3 + CrNa2O4 + CaCO3
• Hg + NaOH = Hg2O + NaH
• Zn + MnO4 = Zn2(MnO4)2
• Recently Balanced Equations

Balance NH2CH2COOH + Na2CO3 = NH2CH2COONa + H2O + CO2 Using the Algebraic Method

To balance the equation NH2CH2COOH + Na2CO3 = NH2CH2COONa + H2O + CO2 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 to represent the unknown coefficients.

a NH2CH2COOH + b Na2CO3 = c NH2CH2COONa + d H2O + f CO2

Step 2: Create a System of Equations

Create an equation for each element (N, H, C, O, Na) where each term represents the number of atoms of the element in each reactant or product.

N: 1a + 0b = 1c + 0d + 0f H: 5a + 0b = 4c + 2d + 0f C: 2a + 1b = 2c + 0d + 1f O: 2a + 3b = 2c + 1d + 2f Na: 0a + 2b = 1c + 0d + 0f

Step 3: Solve For All Variables

Use substitution, Gaussian elimination, or a calculator to solve for each variable.

Using Substitution or Elimination

• 1a - 1c = 0
• 5a - 4c - 2d = 0
• 2a + 1b - 2c - 1f = 0
• 2a + 3b - 2c - 1d - 2f = 0
• 2b - 1c = 0

Using Linear Systems / Algebra

Use your graphing calculator's rref() function (or an online rref calculator) to convert the following matrix into reduced row-echelon-form:

[ 1 0 -1 0 0 0] [ 5 0 -4 -2 0 0] [ 2 1 -2 0 -1 0] [ 2 3 -2 -1 -2 0] [ 0 2 -1 0 0 0] The resulting matrix can be used to determine the coefficients. In the case of a single solution, the last column of the matrix will contain the coefficients. Convert to RREF and Solve Step-by-Step

Simplify the result to get the lowest, whole integer values.

• a = 2 (NH2CH2COOH)
• b = 1 (Na2CO3)
• c = 2 (NH2CH2COONa)
• d = 1 (H2O)
• f = 1 (CO2)

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 NH2CH2COOH + Na2CO3 = 2 NH2CH2COONa + H2O + CO2

Reactants	Products
N	2	2	✔️
H	10	10	✔️
C	5	5	✔️
O	7	7	✔️
Na	2	2	✔️

Since there is an equal number of each element in the reactants and products of 2NH2CH2COOH + Na2CO3 = 2NH2CH2COONa + H2O + CO2, the equation is balanced.

Balance NH2CH2COOH + Na2CO3 = NH2CH2COONa + H2O + CO2 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 end of a chemical reaction as at the beginning. To be balanced, every element in NH2CH2COOH + Na2CO3 = NH2CH2COONa + H2O + CO2 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 to 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
NH2CH2COOH	Na2CO3	Total	NH2CH2COONa	H2O	CO2	Total
N	1	1	1	1	✔️
H	5	5	4	2	6	❌
C	2	1	3	2	1	3	✔️
O	2	3	5	2	1	2	5	✔️
Na	2	2	1	1	❌

Step 2: Multiply coefficients for compounds to balance out each element

For each element that is not equal, try to balance it by adding more of it to the side with less. Sometimes there may be multiple compounds with that element on one side, so you'll need to use your best judgement and be prepared to go back and try the other options.

1. Na is not balanced. Add 1 molecule of NH2CH2COONa to the product (right-hand) side to balance Sodium: NH2CH2COOH + Na2CO3 = 2NH2CH2COONa + H2O + CO2

   Reactants	Products
   N	1	2	❌
   H	5	10	❌
   C	3	5	❌
   O	5	7	❌
   Na	2	2	✔️

2. N is not balanced. Add 1 molecule of NH2CH2COOH to the reactant (left-hand) side to balance Nitrogen: 2NH2CH2COOH + Na2CO3 = 2NH2CH2COONa + H2O + CO2

   Reactants	Products
   N	2	2	✔️
   H	10	10	✔️
   C	5	5	✔️
   O	7	7	✔️
   Na	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.

2NH2CH2COOH + Na2CO3 = 2NH2CH2COONa + H2O + CO2

Practice Balancing

NH2CH2COOH - + + Na2CO3 - + = NH2CH2COONa - + + H2O - + + CO2 - +

Reactants	Products
N	1	1	✔️
H	5	6	❌
C	3	3	✔️
O	5	5	✔️
Na	2	1	❌

Hint: The sum of the balanced coefficients is 7. See Step-By-Step Solution

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