Could you tell me how to balance equations like these?
The numbers with * before them mean that they're subscripts. You can not change the subscripts, but you may put numbers in the blank spaces. The arrow is like an equals sign.
examples-
___C*5H*9O+ ___O*2 --->___CO*2+ ___H*2O
___NH*3+ ___H*2SO*4 --->___(NH*4)*2SO*4
Sure! Balancing chemical equations is an important skill in chemistry. To balance an equation, you need to make sure that the number of atoms of each element on both sides of the equation is the same.
Let's start with the first example:
___C*5H*9O + ___O*2 ---> ___CO*2 + ___H*2O
1. Start by identifying the elements on both sides of the equation. In this case, we have carbon (C), hydrogen (H), and oxygen (O).
2. Begin with the most complex molecule or compound, which in this case is C*5H*9O. Count the number of atoms of each element on both sides of the equation. On the left side: 5 carbon atoms, 9 hydrogen atoms, and 2 oxygen atoms. On the right side: 1 carbon atom, 2 hydrogen atoms, and 2 oxygen atoms.
3. To balance the carbon atoms, we need to put a coefficient (number in front of the compound or element) in front of CO*2. Since there is already one carbon atom on the right side, we put a coefficient of 5 in front of CO*2, making it 5CO*2. Now we have 5 carbon atoms on both sides of the equation.
4. Next, we balance the hydrogen atoms. There are 9 hydrogen atoms on the left side, but only 2 on the right side. To balance them, put a coefficient of 4 in front of H*2O, making it 4H*2O. This gives us 8 hydrogen atoms on both sides.
5. Finally, balance the oxygen atoms. There are 2 oxygen atoms on the left side, but now we have 10 oxygen atoms on the right side (5 multiplied by 2). To balance them, we need to adjust the coefficient of O*2. Dividing both sides by 2 gives us O on the left side and 5O*2 on the right side, which simplifies to 10 oxygen atoms. Therefore, the equation becomes:
5C*5H*9O + O*2 ---> 5CO*2 + 4H*2O
Now, let's move on to the second example:
___NH*3 + ___H*2SO*4 ---> ___(NH*4)*2SO*4
Again, follow the same steps:
1. Identify the elements involved: nitrogen (N), hydrogen (H), and sulfur (S).
2. Start with the most complex compound, H*2SO*4. Count the number of atoms on both sides. On the left side: 1 nitrogen atom, 3 hydrogen atoms, and 4 oxygen atoms. On the right side: 2 nitrogen atoms, 8 hydrogen atoms, and 4 sulfur atoms.
3. To balance the nitrogen atoms, put a coefficient of 2 in front of NH*3, making it 2NH*3. This gives us 2 nitrogen atoms on both sides.
4. Next, balance the hydrogen atoms. There are 3 hydrogen atoms on the left side, but now we have 8 hydrogen atoms on the right side. To balance them, put a coefficient of 4 in front of NH*3, making it 4NH*3. This gives us 12 hydrogen atoms on both sides.
5. Finally, balance the sulfur atoms. There are 1 sulfur atom on the left side, but now we have 4 sulfur atoms on the right side. To balance them, adjust the coefficient of (NH*4)*2SO*4 by putting a coefficient of 2 in front of NH*4. This gives us 2 sulfur atoms on both sides.
The balanced equation is:
2NH*3 + 4H*2SO*4 ---> (NH*4)*2SO*4
I hope this explanation helps you understand how to balance chemical equations. Remember to always check if the equation is balanced by counting the number of atoms on both sides.