Balance the following equation. (for a balanced eq. aA + bB → cC + dD, enter your answer as the integer abcd)
MnO4−(aq) + SO32−(aq) + H+(aq) → Mn2+(aq) + SO42−(aq) + H2O(l)
i got 2525 and its going in as incorrect
Now you get to balance this equation (answer in the same way as in the problem above):
Al(s) + NO3−(aq) + OH−(aq) + H2O → Al(OH)4−(aq) + NH3(g)
i go 2121 but its not correct
MnO4−(aq) + SO32−(aq) + H+(aq) → Mn2+(aq) + SO42−(aq) + H2O(l)
in the form
a*MnO4−(aq) + b*SO32−(aq) + c*H+(aq) → d*Mn2+(aq) + e*SO42−(aq) + f*H2O(l)
Start by counting how many of each are on each side. I'm going to choose to use the convention [element (left, right)]
Mn(1,1)
O(7,5)
S(1,1)
H(1,2)
b=e; (Balancing the sulfur, S)
a=d; (Balancing the Mn)
c=2*f; (Balancing the hydrogen)
3*b + 4*a = 4*e + f (Balancing the oxygen)
So let a=e = 1; b=1; f=3
(3*1 + 4*1 = 4*1 + 3)
d=1
c=6
So the equation is
MnO4−(aq) + SO32−(aq) + 6H+(aq) → Mn2+(aq) + SO42−(aq) + 3H2O(l)
2nd problem:
Al(s) + NO3−(aq) + OH−(aq) + H2O → Al(OH)4−(aq) + NH3(g)
of the form
a*Al(s) + b*NO3−(aq) + c*OH−(aq) + d*H2O → e*Al(OH)4−(aq) + f*NH3(g)
Al(1,1)
N(1,1)
O(5,4)
H(3,7)
a=e (balancing the aluminum)
b=f (balancing the nitrogen)
3*b + c + d = 4*e (balancing the oxygen)
c + 2*d = 4*e + 3*f (Balancing the hydrogen)
You should be able to experiment with numbers and balance this out
To balance the equation Al(s) + NO3−(aq) + OH−(aq) + H2O → Al(OH)4−(aq) + NH3(g), let's go step by step:
1. Start by balancing the aluminum (Al) atoms on both sides.
Al(s) + NO3−(aq) + OH−(aq) + H2O → Al(OH)4−(aq) + NH3(g)
2. Next, balance the nitrogen (N) atoms. On the left side, there is one nitrate ion (NO3−), and on the right side, there is one ammonia (NH3) molecule. We can add a coefficient of 2 in front of NH3 to balance the N atoms.
Al(s) + NO3−(aq) + OH−(aq) + H2O → Al(OH)4−(aq) + 2 NH3(g)
3. Now let's balance the hydrogen (H) atoms. On the left side, there are already two H atoms from the water molecule (H2O). On the right side, there are two OH− ions and one Al(OH)4−, which together contribute 8 H atoms. We can add a coefficient of 4 in front of the water molecule to balance the H atoms.
Al(s) + NO3−(aq) + OH−(aq) + 4 H2O → Al(OH)4−(aq) + 2 NH3(g)
4. Finally, let's balance the oxygen (O) atoms. On the left side, there are 4 oxygen atoms from the four water molecules (4H2O). On the right side, there are 4 oxygen atoms in the Al(OH)4− ion. We can see that the NO3− ion will contribute 3 additional oxygen atoms. Therefore, we can balance the oxygen atoms by adding a coefficient of 3 in front of the nitrate ion (NO3−).
Al(s) + 3 NO3−(aq) + OH−(aq) + 4 H2O → Al(OH)4−(aq) + 2 NH3(g)
Now the equation is balanced, and the coefficients are:
a = 1
b = 3
c = 1
d = 2
So, the balanced equation can be represented as: Al(s) + 3 NO3−(aq) + OH−(aq) + 4 H2O → Al(OH)4−(aq) + 2 NH3(g).
Therefore, the balanced equation can be written as 1304.
To balance a chemical equation, you need to ensure that the number of atoms on both sides of the equation are equal. To do this, you can follow the steps below:
For the first equation: MnO4−(aq) + SO32−(aq) + H+(aq) → Mn2+(aq) + SO42−(aq) + H2O(l)
Step 1: Count the number of atoms of each element on both sides of the equation:
On the left side: Mn: 1, O: 4, S: 1, H: 1.
On the right side: Mn: 1, O: 4, S: 1, H: 2.
Step 2: Start by balancing the elements that appear in the least number of compounds or are in the most complex compounds. In this case, we start with H.
Step 3: To balance the H atoms, you can add a coefficient of 2 in front of the H+ on the left side:
MnO4−(aq) + SO32−(aq) + 2H+(aq) → Mn2+(aq) + SO42−(aq) + H2O(l)
Step 4: Now, let's balance the O atoms. On the left side, there are 4 O atoms from the MnO4− and 3 O atoms from the SO32−, totaling 7 O atoms. On the right side, there are only 4 O atoms from the SO42−. To balance this, we can add a coefficient of 3/2 (or 1.5) in front of the SO42− on the right side:
MnO4−(aq) + SO32−(aq) + 2H+(aq) → Mn2+(aq) + 3/2SO42−(aq) + H2O(l)
Step 5: Multiply all the coefficients by 2 to eliminate fractions and obtain a whole-number balanced equation:
2MnO4−(aq) + 3SO32−(aq) + 4H+(aq) → 2Mn2+(aq) + 3SO42−(aq) + 2H2O(l)
Therefore, the balanced equation is: 2MnO4−(aq) + 3SO32−(aq) + 4H+(aq) → 2Mn2+(aq) + 3SO42−(aq) + 2H2O(l).
Now, let's move on to balancing the second equation:
Al(s) + NO3−(aq) + OH−(aq) + H2O → Al(OH)4−(aq) + NH3(g)
Step 1: Count the number of atoms of each element on both sides of the equation:
On the left side: Al: 1, N: 1, O: 3, H: 3
On the right side: Al: 1, N: 1, O: 5, H: 1.
Step 2: Start by balancing the elements that appear in the least number of compounds or are in the most complex compounds. In this case, we start with Al.
Step 3: To balance the Al atom, you can add a coefficient of 1 in front of the Al on the right side:
Al(s) + NO3−(aq) + OH−(aq) + H2O → Al(OH)4−(aq) + NH3(g)
Step 4: Now, let's balance the H atoms. On the left side, there are 3 H atoms from the OH− and 3 H atoms from the H2O, totaling 6 H atoms. On the right side, there is only 1 H atom from the NH3. To balance this, we can add a coefficient of 6 in front of the NH3 on the right side:
Al(s) + NO3−(aq) + OH−(aq) + 6H2O → Al(OH)4−(aq) + 6NH3(g)
Step 5: Finally, let's balance the O atoms. On the left side, there are 3 O atoms from the NO3− and 3 O atoms from the OH−, totaling 6 O atoms. On the right side, there are 12 O atoms from the Al(OH)4− and 0 O atoms from the NH3. To balance this, we can add a coefficient of 6 in front of the Al(OH)4− on the right side:
Al(s) + NO3−(aq) + OH−(aq) + 6H2O → 6Al(OH)4−(aq) + 6NH3(g)
Therefore, the balanced equation is: Al(s) + NO3−(aq) + OH−(aq) + 6H2O → 6Al(OH)4−(aq) + 6NH3(g).
As a result, the answer as the integer representation of abcd is 6036.