balance the following chemical equations.\n1. fe + h₂so₄ → fe₂(so₄)₃ + h₂\n2. c₂h₆ + o₂ → h₂o + co₂\n3. koh…

Explanation:

Step1: Balance Fe in equation 1

For the equation $Fe + H_2SO_4\rightarrow Fe_2(SO_4)_3 + H_2$, to balance the Fe atoms, since there are 2 Fe atoms in $Fe_2(SO_4)_3$, we put 2 in front of Fe: $2Fe + H_2SO_4\rightarrow Fe_2(SO_4)_3 + H_2$. Then to balance the $SO_4$ groups, we put 3 in front of $H_2SO_4$, and the balanced equation is $2Fe + 3H_2SO_4\rightarrow Fe_2(SO_4)_3+3H_2$.

Step2: Balance C and H in equation 2

For $C_2H_6+O_2\rightarrow H_2O + CO_2$, first balance C. Since there are 2 C - atoms in $C_2H_6$, we put 2 in front of $CO_2$. Then balance H. There are 6 H - atoms in $C_2H_6$, so we put 3 in front of $H_2O$. Now, for O - atoms, on the right - hand side we have $2\times2 + 3\times1=7$ O - atoms. So we put $\frac{7}{2}$ in front of $O_2$. To get whole numbers, we multiply all coefficients by 2, and the balanced equation is $2C_2H_6 + 7O_2\rightarrow6H_2O + 4CO_2$.

Step3: Balance K and P in equation 3

For $KOH+H_3PO_4\rightarrow K_3PO_4 + H_2O$, to balance K, we put 3 in front of KOH. Then the equation becomes $3KOH + H_3PO_4\rightarrow K_3PO_4+3H_2O$.

Step4: Balance Sn and O in equation 4

For $SnO_2+H_2\rightarrow Sn + H_2O$, there is 1 Sn on both sides. To balance O, since there are 2 O - atoms in $SnO_2$, we put 2 in front of $H_2O$. Then to balance H, we put 2 in front of $H_2$. The balanced equation is $SnO_2 + 2H_2\rightarrow Sn+2H_2O$.

Step5: Balance N and H in equation 5

For $NH_3+O_2\rightarrow NO + H_2O$, first balance N, which is already 1 on both sides. Then balance H. Since there are 3 H - atoms in $NH_3$, we put $\frac{3}{2}$ in front of $H_2O$. For O, on the right - hand side we have $1 + \frac{3}{2}\times1=\frac{5}{2}$ O - atoms. So we put $\frac{5}{4}$ in front of $O_2$. To get whole numbers, we multiply all coefficients by 4, and the balanced equation is $4NH_3+5O_2\rightarrow4NO + 6H_2O$.

Step6: Balance K and N in equation 6

For $KNO_3+H_2CO_3\rightarrow K_2CO_3 + HNO_3$, to balance K, we put 2 in front of $KNO_3$. Then the balanced equation is $2KNO_3+H_2CO_3\rightarrow K_2CO_3 + 2HNO_3$.

Step7: Balance B and Br in equation 7

For $B_2Br_6+HNO_3\rightarrow B(NO_3)_3 + HBr$, first balance B. Since there are 2 B - atoms in $B_2Br_6$, we put 2 in front of $B(NO_3)_3$. Then to balance Br, we put 6 in front of $HBr$. Now, to balance N, we put 6 in front of $HNO_3$. The balanced equation is $B_2Br_6+6HNO_3\rightarrow2B(NO_3)_3 + 6HBr$.

Step8: Balance B and Li in equation 8

For $BF_3+Li_2SO_3\rightarrow B_2(SO_3)_3 + LiF$, first balance B. We put 2 in front of $BF_3$. Then to balance Li, we put 3 in front of $Li_2SO_3$. The balanced equation is $2BF_3+3Li_2SO_3\rightarrow B_2(SO_3)_3 + 6LiF$.

Step9: Balance Pb and P in equation 9

For $(NH_4)_3PO_4+Pb(NO_3)_4\rightarrow Pb_3(PO_4)_4+NH_4NO_3$, first balance Pb. We put 3 in front of $Pb(NO_3)_4$. Then to balance P, we put 4 in front of $(NH_4)_3PO_4$. Now, to balance N and H, we put 12 in front of $NH_4NO_3$. The balanced equation is $4(NH_4)_3PO_4+3Pb(NO_3)_4\rightarrow Pb_3(PO_4)_4 + 12NH_4NO_3$.

Step10: Balance Se and Cl in equation 10

For $SeCl_6+O_2\rightarrow SeO_2 + Cl_2$, first balance Se, which is 1 on both sides. Then to balance Cl, since there are 6 Cl - atoms in $SeCl_6$, we put 3 in front of $Cl_2$. Now, to balance O, we put 1 in front of $O_2$ and 1 in front of $SeO_2$. The balanced equation is $SeCl_6+O_2\rightarrow SeO_2 + 3Cl_2$.

Answer:

1. $2Fe + 3H_2SO_4\rightarrow Fe_2(SO_4)_3+3H_2$
2. $2C_2H_6 + 7O_2\rightarrow6H_2O + 4CO_2$
3. $3KOH + H_3PO_4\rightarrow K_3PO_4+3H_2O$
4. $SnO_2 + 2H_2\rightarrow Sn+2H_2O$
5. $4NH_3+5O_2\rightarrow4NO + 6H_2O$
6. $2KNO_3+H_2CO_3\rightarrow K_2CO_3 + 2HNO_3$
7. $B_2Br_6+6HNO_3\rightarrow2B(NO_3)_3 + 6HBr$
8. $2BF_3+3Li_2SO_3\rightarrow B_2(SO_3)_3 + 6LiF$
9. $4(NH_4)_3PO_4+3Pb(NO_3)_4\rightarrow Pb_3(PO_4)_4 + 12NH_4NO_3$
10. $SeCl_6+O_2\rightarrow SeO_2 + 3Cl_2$