Question

listed in the item bank are some important labels for sections of the image below. drag and drop each label to the corresponding area it identifies in the image. place the number in the correct coefficient location to make each equation balance. 2 c₂h₆ + 7o₂ → 6h₂o + 4co₂ hclo₄ + p₄o₁₀ → 4h₃po₄ + cl₂o₇ 2n₂ (g) + o₂(g) + 2h₂o → 4hno₃(aq) cacl₂ + 2na₃po₄ → ca₃(po₄)₂ + 6nacl fes + o₂ → 2fe₂o₃ + 4so₂

Explanation:

Step1: Balance first equation

For $C_2H_6 + 7O_2
ightarrow6H_2O + 4CO_2$, the coefficient of $C_2H_6$ is 2 as given.

Step2: Balance second equation

For $HClO_4+P_4O_{10}
ightarrow4H_3PO_4 + Cl_2O_7$, balance P first. Since there are 4 P in $P_4O_{10}$, we need 4 moles of $H_3PO_4$. Then balance Cl and O. The balanced equation is $12HClO_4 + P_4O_{10}
ightarrow4H_3PO_4+6Cl_2O_7$, so the coefficients are 12 and 6.

Step3: Balance third equation

For $2N_2+O_2 + 2H_2O
ightarrow4HNO_3$, balance N first. There are 4 N in $4HNO_3$ and 2 moles of $N_2$. Then balance O. The coefficient of $O_2$ is 5.

Step4: Balance fourth equation

For $CaCl_2+2Na_3PO_4
ightarrow Ca_3(PO_4)_2+6NaCl$, balance Ca first. We need 3 moles of $CaCl_2$.

Step5: Balance fifth equation

For $FeS+O_2
ightarrow2Fe_2O_3 + 4SO_2$, balance Fe first. We need 4 moles of $FeS$. Then balance S and O. The coefficient of $O_2$ is 7.

Answer:

First equation: 2
Second equation: 12, 6
Third equation: 5
Fourth equation: 3
Fifth equation: 4, 7