Need help please
Which of the following equations has the coefficients 2,1,1,2 when it is balanced?
Fe2O3 + HClO4 → Fe(ClO4)3 + H2O
Ca(OH)2 + H3PO4 → Ca3(PO4)2 + H2O
KOH + H2SO4 → K2SO4 + H2O
Ba(OH)2 + P4O10 → Ba3(PO4)2 + H2O
Al2O3 + H2SeO4 → Al2(SeO4)3 + H2O
Which is the only equation with no 3s in it? It can't be 1, 2, 4, or 5 because all of those require a three somewhere to balance it. And the answer has no other 3s.
Ba(OH)2 + P4O10 → Ba3(PO4)2 + H2O
To determine which equation has the coefficients 2, 1, 1, 2 when balanced, we need to balance each equation and see which one matches the given coefficients.
Let's go through each equation one by one:
1. Fe2O3 + HClO4 → Fe(ClO4)3 + H2O
To balance this equation, we need to ensure that the number of atoms on each side of the equation is equal. Let's start with oxygen (O).
On the left-hand side, we have 3 oxygen atoms from HClO4, and on the right-hand side, we have 3 oxygen atoms from Fe(ClO4)3 and 1 oxygen atom from H2O. In total, there are 4 oxygen atoms on the right-hand side. So, we need to balance oxygen by multiplying H2O by 4:
Fe2O3 + HClO4 → Fe(ClO4)3 + 4H2O
Now, let's balance hydrogen (H). On the left-hand side, we have 4 hydrogen atoms from HClO4, and on the right-hand side, we have 8 hydrogen atoms from 4H2O. So, hydrogen is already balanced.
Next, let's balance chlorine (Cl). On the left-hand side, we have 1 chlorine atom from HClO4, and on the right-hand side, we have 3 chlorine atoms from Fe(ClO4)3. To balance chlorine, we need to multiply HClO4 by 3:
Fe2O3 + 3HClO4 → Fe(ClO4)3 + 4H2O
Finally, let's balance iron (Fe). On the left-hand side, we have 2 iron atoms, and on the right-hand side, we have 1 iron atom. To balance iron, we need to multiply Fe(ClO4)3 by 2:
Fe2O3 + 3HClO4 → 2Fe(ClO4)3 + 4H2O
Therefore, the coefficients in this equation are 2, 3, 2, and 4. It does not match the given coefficients.
Now, let's move to the next equation:
2. Ca(OH)2 + H3PO4 → Ca3(PO4)2 + H2O
Again, let's balance oxygen first. On the left-hand side, we have 2 oxygen atoms from Ca(OH)2, and on the right-hand side, we have 8 oxygen atoms from Ca3(PO4)2 and 1 oxygen atom from H2O. In total, there are 9 oxygen atoms on the right-hand side. So, we need to balance oxygen by multiplying Ca(OH)2 by 4:
4Ca(OH)2 + H3PO4 → Ca3(PO4)2 + 4H2O
Next, let's balance hydrogen. On the left-hand side, we have 8 hydrogen atoms from Ca(OH)2, and on the right-hand side, we have 3 hydrogen atoms from H3PO4 and 8 hydrogen atoms from 4H2O. In total, there are 11 hydrogen atoms on the right-hand side. So, we need to balance hydrogen by multiplying H3PO4 by 8:
4Ca(OH)2 + 8H3PO4 → Ca3(PO4)2 + 4H2O
Now, let's balance calcium (Ca) and phosphorus (P). On the left-hand side, we have 4 calcium atoms and 8 phosphorus atoms. On the right-hand side, we have 3 calcium atoms from Ca3(PO4)2. To balance calcium, we need to multiply Ca(OH)2 by 3, and to balance phosphorus, we need to multiply H3PO4 by 2:
12Ca(OH)2 + 16H3PO4 → 3Ca3(PO4)2 + 4H2O
Therefore, the coefficients in this equation are 12, 16, 3, and 4. It does not match the given coefficients.
Let's move on to the next equation:
3. KOH + H2SO4 → K2SO4 + H2O
First, let's balance hydrogen (H) and sulfur (S). On the left-hand side, we have 2 hydrogen atoms from H2SO4, and on the right-hand side, we have 2 hydrogen atoms from H2O. Both hydrogen and sulfur are already balanced.
Next, let's balance potassium (K) and oxygen (O). On the left-hand side, we have 1 potassium atom and 1 oxygen atom from KOH. On the right-hand side, we have 2 potassium atoms from K2SO4. To balance potassium, we need to multiply KOH by 2:
2KOH + H2SO4 → K2SO4 + H2O
Therefore, the coefficients in this equation are 2, 1, 1, and 1. It matches the given coefficients.
So, the equation with the coefficients 2, 1, 1, 2 when balanced is:
KOH + H2SO4 → K2SO4 + H2O