Question

for the chemical equation below, list the stoichiometric coefficients in the order they would appear in the equation (1s included)
__ca(oh)₂ + cu₃po₄ ==> ca₃(po₄)₂ + __cuoh

○ 6:2:1:6
○ 3:1:1:3
○ 3:2:1:3
○ 3:2:1:6

Explanation:

Step1: Balance Ca atoms

Left: 1 Ca (in \( \text{Ca(OH)}_2 \)), Right: 3 Ca (in \( \text{Ca}_3(\text{PO}_4)_2 \)). So coefficient of \( \text{Ca(OH)}_2 \) is 3 (to make 3 Ca on left). Now equation: \( 3\text{Ca(OH)}_2 + \text{Cu}_3\text{PO}_4
ightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{CuOH} \)

Step2: Balance PO₄³⁻ ions

Left: 1 \( \text{PO}_4^{3-} \) (in \( \text{Cu}_3\text{PO}_4 \)), Right: 2 \( \text{PO}_4^{3-} \) (in \( \text{Ca}_3(\text{PO}_4)_2 \)). So coefficient of \( \text{Cu}_3\text{PO}_4 \) is 2 (to make 2 \( \text{PO}_4^{3-} \) on left). Now: \( 3\text{Ca(OH)}_2 + 2\text{Cu}_3\text{PO}_4
ightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{CuOH} \)

Step3: Balance Cu atoms

Left: 6 Cu (2×3 in \( 2\text{Cu}_3\text{PO}_4 \)), Right: 1 Cu (in \( \text{CuOH} \)). So coefficient of \( \text{CuOH} \) is 6 (to make 6 Cu on right). Now: \( 3\text{Ca(OH)}_2 + 2\text{Cu}_3\text{PO}_4
ightarrow \text{Ca}_3(\text{PO}_4)_2 + 6\text{CuOH} \)

Step4: Check OH⁻ (or O, H)

Left: 3×2 = 6 OH⁻, Right: 6×1 = 6 OH⁻. Balanced. So coefficients: 3, 2, 1, 6.

Answer:

3:2:1:6 (corresponding to the option "3:2:1:6")