how do you balance the equation with H3PO4(aq)and BaCl2(aq)?
Sam/Sally/??? Did you get this far? By the way you would have obtained a faster answer if you had asked at the original question post.
BaCl2 + H3PO4 ==> Ba3(PO4)2 + H2O
To balance the chemical equation with H3PO4(aq) and BaCl2(aq), follow these steps:
Step 1: Write down the unbalanced equation:
H3PO4(aq) + BaCl2(aq) -> ?
Step 2: Count the number of each type of atom on both sides of the equation. In this case, we have:
On the left side: 3 hydrogen atoms (H), 1 phosphorus atom (P), 4 oxygen atoms (O), 1 barium atom (Ba), and 2 chlorine atoms (Cl).
On the right side: ? (we do not know the coefficients yet).
Step 3: Start by balancing the elements that occur in the least number of compounds. In this case, phosphorus appears in only one compound on the left side, so let's balance that first.
H3PO4(aq) + BaCl2(aq) -> ?H3PO4(aq) + ?BaCl2(aq)
Step 4: To balance phosphorus, place a coefficient of 3 in front of H3PO4 on the left side:
3H3PO4(aq) + BaCl2(aq) -> ?H3PO4(aq) + ?BaCl2(aq)
Step 5: Now, let's balance the chloride atoms (Cl). There are 2 chloride atoms in BaCl2 and only 1 on the right side, so we need to balance that.
Place a coefficient of 2 in front of BaCl2 on the right side:
3H3PO4(aq) + BaCl2(aq) -> ?H3PO4(aq) + 2BaCl2(aq)
Step 6: Next, let's balance the hydrogen atoms (H). There are now 9 hydrogen atoms on the left side and 6 hydrogen atoms on the right side.
To balance hydrogen, place a coefficient of 6 in front of H3PO4 on the right side:
3H3PO4(aq) + BaCl2(aq) -> 6H3PO4(aq) + 2BaCl2(aq)
Step 7: Finally, let's balance the barium atoms (Ba). There are 2 barium atoms on the right side and only 1 on the left side.
To balance barium, place a coefficient of 2 in front of BaCl2 on the left side:
3H3PO4(aq) + 2BaCl2(aq) -> 6H3PO4(aq) + 2BaCl2(aq)
Now the equation is balanced with 3H3PO4(aq) + 2BaCl2(aq) on the left side.