This model will teach you how to balance the chemical equations. Balancing chemical equations is necessary in order to determine (on the basis of the law of conservation of matter) the proportions of reagents that are required to complete a chemical reaction, as well as the amount of the reaction's product. A chemical equation is considered balanced if the numbers of each type of atoms on the left and on the right sides of the equation are equal. Select a chemical equation for balancing. To do this, choose the desired equation from the scrollable list of reactions located below the task window. The chosen equation will appear in the task window. Choose a term with the changing factor from the list that is located below the list of equations. The coefficient value is to be set using the input field with scrolling buttons, located to the right from the list of equation members. Upon completion of balancing the equation, you can make sure that the calculation is correct by pressing "Check" button. The results of this control process will appear in the small window that is located between the task window and the list of reactions. Let us analyze the process of chemical equation balancing with the equation **BF**_{3} + NaBH_{4} = NaBF_{4} + B_{2}H_{6 }as an example. Consider the atoms of fluorine. In order to make their numbers on the right and the left sides of equation equal, the **BF**_{3} and **NaBF**_{4} coefficient ratio must equal to 4/3 (∙). Now, let us consider the atoms of hydrogen. In order to make their numbers satisfy both sides of the equation, **NaBH4 **and **B2H6** coefficient ratio should be equal to 3/2 (∙∙). Remember that the coefficients of **NaBH4** on both sides of equation should be equal (∙∙∙), in order to equalize the numbers of sodium atoms in the reagents and products. So we can assume that the coefficient of **BF**_{3} will be equal to 4, and according to condition (∙), the coefficient of **NaBF**_{4} will be 3. Condition (∙∙∙) results in the coefficient of **NaBH**_{4} being equal to 3 as well, and that, according to condition (∙∙), leaves **B**_{2}H_{6} with a coefficient of 2. Note that since the coefficients of all products and reagents have been already determined, the numbers of boron atoms on the left and the right sides of the equation have been automatically equalized. Let us now look at another example. Consider a redox reaction between **KMnO**_{4} - HCl (Our model does not contain this problem). Input data: **MnO**_{4}^{-} is reduced to **Mn**^{2+} by adding 5 electrons in acid medium, and **Cl**^{-} undergoes oxidization as a reducing agent - to free up chlorine **Cl**_{2} - by losing one electron. Write down the equations of the two components of the reaction in question: reduction: **MnO**_{4}^{-} + 5a + 8 H^{+} = Mn^{2+} + 4 H_{2}O oxidization: **2 Cl**^{-} - 2e = Cl_{2} These two chemical equations (semi-reaction terms of the larger equation) are balanced in accordance with the same principles as the ones that are introduced in the first example. Remember that the total electric charge values on both sides of the semi-reactions must be the same. The total number of electrons (both added and lost) must be equal to 10 (which is the least common multiple of the numbers of electrons added in one semi-reaction and those given up in the other semi-reaction). The following equation can be obtained by multiplying the reduction and oxidization semi-reactions respectively by 2 and 5, and adding these semi-reactions together: **2 MnO**_{4}^{-} + 16 H^{+} + 10 Cl^{-} = 2 Mn^{2+} + 5 Cl_{2} + 8 H_{2}O **
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