Balance the equation C2H6 (g) + O2 (g) ---> CO2 (g) + H2O (l)?
Balanced equation: C2H6 + 4O2 --> 2CO2 + 6H2O 6x32= 132 moles of H20
How to balance the equation?
Okay, this is a simple acid/bace equation. The "H" from the Acetic Acid will combine with the "OH" from your base to produce H20. You can use the formula for double displacement (Ax +By -> Bx + Ay) to combine the Acetate with the Barium. So, you have: HC2H3O2 + Ba(OH)2 -> H2O + Ba(C2H3O2)2 Since Barium has a charge of +2 and Acetate has a charge of -1, you need 2 Acetate compounds to balance their charge. Now, you need to count the elements on each side of the equation, to make sure they are balanced. There are 6 H atoms on the left side, but there are 8 H atoms on the right side. There are 2 atoms of C on the left side of the equation, but there are 4 on the right. There are 4 atoms of O on the left side, but there are 5 on the right side. And there are 2 atoms of Ba on both sides. So, the H, C, and O need to be balanced. By placing coefficients in front of compounds in the formula, you can easily balance the equation. Balanced equation: 2 HC2H3O2 + Ba(OH)2 -> 2 H20 + Ba(C2H3O2)2 To check, again count the number of elements on each side of the equation. There are 10 H atoms on both sides, 4 C atoms on both sides, 6 O atoms on both sides, and 2 Ba atoms on both sides. The chemical equation is balanced. (:
How do you balance this equation: H2S + O2 = H2O + SO2?
H2S + O2 → H2O + SO2Number of each atom on the reactant side: H is 2; S is 1; O is 2Number of each atom on the product side: H is 2; S is 1; O is 3Therefore, oxygen is out of balance. The common factor for 2 and 3 is 6, hence make oxygen 6 on the reactant side with a coefficient of 3 means the other atoms are out of balance as shown below:H2S + 3O2 → H2O + SO2Reactant side: H is 2; S is 1; O is 6Product side: H is 2; S is 1; O is 3To start with, you can place a coefficient of 2 in front of H2O to balance out the oxygen, giving:H2S + 3O2 → 2H2O + SO2Reactant side: H is 2; S is 1; O is 6Product side: H is 4; S is 1; O is 4Hydrogen is now out of balance, so a coefficient of 2 can be placed in front of H2S to give:2H2S + 3O2 → 2H2O + SO2Reactant side: H is 4; S is 2; O is 6Product side: H is 4; S is 1; O is 4Sulphur is still out of balance, so a coefficient of 2 can be placed in front of SO2 to give:2H2S + 3O2 → 2H2O + 2SO2Reactant side: H is 4; S is 2; O is 6Product side: H is 4; S is 2; O is 6I hope that helps
How do you balance this equation? CO2 + H2O → C6H12O6 + O2?
I would start with the Carbon. It's really a toss-up between it and the Hydrogen, but you can just ignore the Oxygen off the top, since it is by itself on one side, making it easy to fix at the end. Since you have minimum 6 Carbons on the right, you need at least 6 CO2 to balance the Carbon, as CO2 has only 1 each. 6CO2 + H2O -----> C6H12O6 + O2 The Carbon is balanced, so let's try to balance the Hydrogen now. Since there are 12 Hydrogens on the right, we need 6 H2O to balance, as there are 2 Hydrogens in each. 6CO2 + 6H2O -----> C6H12O6 + O2 Checking both sides again, we see that both sides have the same number of Carbon and Hydrogen, so now let's look at the Oxygen. Counting the left side, we see that there are 6(2) + 6 = 18 Oxygen. However, the right side only has 6 + 2 = 8 Oxygen. Fortunately, the difference of 10 is an even number, so all we need to do is increase the coefficient for O2 until we get 18 on the right side. Since each additional O2 provides 2 additional Oxygen, we will need 10/2=5 ADDITIONAL O2. The new coefficient should be 6, giving: 6CO2+6H2O -----> C6H12O6 + 6O2 Checking both sides, we now see that it is balanced.
How can you balance the equation for NH3?
N2 +3H2 -> 2NH3Both the reactant and the product side of the chemical equation have to mathematically balance.https://twitter.com/pati_pandi
How do I balance this equation: CO2 + H2O = C6H12O6 +O2 + H2O?
This is the equation for photosynthesis whereby, in the presence of light energy from the Sun and chlorophyll, plants synthesise glucose from carbon dioxide and water. As water is present as a reactant, it doesn’t also appear on the ’products’ side of the equation. So:Water + carbon dioxide > glucose + oxygen.As glucose contains 6 carbon atoms, these must be obtained from 6 molecules of CO2.Similarly, glucose contains 12 atoms of hydrogen. These must be obtained from half that number of molecules of H2O. Finally, one must have sufficient oxygen molecules produced to balance the number of oxygen atoms present in the reactants. So, overall:6CO2(g) + 6H2O(l) > C6H12O6(aq) + 6O2(g).
How do you balance this equation H2S+SO2 = S8+H2O?
Try to write it this way:a H2S + b SO2 = c S8 + d H2OYou will then see that8c = a+ bbecause the total of S must be equal on both sides.Also 2a = 2d; so a=d to match the number of HAnd 2b = d to match the number of O.We are left with a linear system of 3 equations on 4 unknowns. Obviously there are infinitely many solutions but we are looking for the smallest positive one.a+b=8ca=2bSo 8c=3bThis shows that c must be divisible by 3 and b must be divisible by 8.You can then confirm that a=16, b=8,c=3,d=16 works.
How do you balance the equation HNO3 + H2S ---> NO + S + H2O?
I think this will solve your problem 2(HNO3) + 3(H2S) -->2(NO) + 3S +4 (H2O)
How to balance the equation of C4H9OH + O2 >CO2 + H2O ?
The point of balancing an equation is to get the same number of each atom on each side. What I usually do (especially when there is water and oxygen) is balance the non oxygen and hydrogen atoms first. So you have 4 Carbon on the left side, and only one on the right. So you multiply the molecule containing Carbon on the right (CO2) by 4 to balance the carbons. Now you equation looks like this: C4H9OH + O2 --> 4 CO2 + H2O Now count the number of Hydrogens. You have 10 on the left side and 2 on the right side. Because the molecule containing the Hydrogen on the right has 2 hydrogen atoms, you only have to multiply by 5. Now your equation looks like this: C4H9OH + O2 --> 4 CO2 + 5 H2O Now count the Oxygens. You have 3 on the left side, and 13 on the right side. You need 10 more on the left side. So add 5 more molecules of O2, giving you 6 in all. Your balanced equation: C4H9OH + 6 O2 --> 4 CO2 + 5 H2O Now you can go back through and make sure you have the same number of each on each side. This is a good way to approach all equations. If you have more types of molecules, I would suggest starting with them, nad ending with balancing the C, O, and H. Remember, you can only change the big numbers out front, not the small numbers in the middle!