Question

Расставьте пропущенные коэффициенты в записях реакций горения: a) 2H2S + O2 = SO2 + H2O; б) 4NH3 + O2 = N2 + H2O; в) 2СН30Н + O2 = CO2 + H2O.

Answer 1

a) 2H2S + 3O2 = 2SO2 + 2H2O b) 4NH3 + 5O2 = 4N2 + 6H2O c) 2СH3OH + 3O2 = 2CO2 + 4H2O

In the given reactions, we need to balance the coefficients to ensure that the number of atoms of each element is the same on both sides of the equation. Let's go through each reaction and balance the coefficients:

a) 2H2S + 3O2 = 2SO2 + 2H2O To balance the equation, we need to ensure that the number of sulfur (S) atoms and hydrogen (H) atoms are the same on both sides. We can start by balancing the sulfur atoms: On the left side, we have 2 sulfur atoms from H2S, and on the right side, we have 2 sulfur atoms from SO2. So, the sulfur atoms are already balanced. Next, let's balance the hydrogen atoms: On the left side, we have 4 hydrogen atoms from 2H2S, and on the right side, we have 4 hydrogen atoms from 2H2O. So, the hydrogen atoms are also balanced. Finally, let's balance the oxygen atoms: On the left side, we have 6 oxygen atoms from 3O2, and on the right side, we have 4 oxygen atoms from 2SO2 and 2 oxygen atoms from 2H2O. To balance the oxygen atoms, we need to multiply the O2 on the left side by 2: 2H2S + 6O2 = 2SO2 + 2H2O

b) 4NH3 + 5O2 = 4N2 + 6H2O To balance the equation, we need to ensure that the number of nitrogen (N) atoms and hydrogen (H) atoms are the same on both sides. We can start by balancing the nitrogen atoms: On the left side, we have 4 nitrogen atoms from 4NH3, and on the right side, we have 4 nitrogen atoms from 4N2. So, the nitrogen atoms are already balanced. Next, let's balance the hydrogen atoms: On the left side, we have 12 hydrogen atoms from 4NH3, and on the right side, we have 12 hydrogen atoms from 6H2O. So, the hydrogen atoms are also balanced. Finally, let's balance the oxygen atoms: On the left side, we have 10 oxygen atoms from 5O2, and on the right side, we have 12 oxygen atoms from 6H2O. To balance the oxygen atoms, we need to multiply the O2 on the left side by 6: 4NH3 + 6O2 = 4N2 + 6H2O

c) 2СH3OH + 3O2 = 2CO2 + 4H2O To balance the equation, we need to ensure that the number of carbon (C) atoms and hydrogen (H) atoms are the same on both sides. We can start by balancing the carbon atoms: On the left side, we have 2 carbon atoms from 2CH3OH, and on the right side, we have 2 carbon atoms from 2CO2. So, the carbon atoms are already balanced. Next, let's balance the hydrogen atoms: On the left side, we have 6 hydrogen atoms from 2CH3OH, and on the right side, we have 8 hydrogen atoms from 4H2O. To balance the hydrogen atoms, we need to multiply the CH3OH on the left side by 4: 8CH3OH + 3O2 = 8CO2 + 16H2O Finally, let's balance the oxygen atoms: On the left side, we have 24 oxygen atoms from 3O2, and on the right side, we have 32 oxygen atoms from 16H2O and 8 oxygen atoms from 8CO2. To balance the oxygen atoms, we need to multiply the O2 on the left side by 8: 8CH3OH + 8O2 = 8CO2 + 16H2O

So, the balanced equations are: a) 2H2S + 6O2 = 2SO2 + 2H2O b) 4NH3 + 6O2 = 4N2 + 6H2O c) 8CH3OH + 8O2 = 8CO2 + 16H2O

Please let me know if you need any further assistance!