Question

the law of conservation of matter tells us that the total number of atoms does not change in a chemical reaction and thus mass is conserved. select all of the reactions that obey the law of conservation of matter
a 4 nh₃ + 5 o₂ → 4 no + 6 h₂o
b 8 h₂s + 8 cl₂ → s₈ + 16 hcl
c ch₄ + 2 o₂ → co₂ + h₂o
d hcl + k₂co₃ → kcl + h₂o + co₂
e 2 al(oh)₃ + 3 h₂so₄ → al₂(so₄)₃ + 6 h₂o

Explanation:

Step1: Check atoms in reaction A

For $4NH_3 + 5O_2
ightarrow4NO+6H_2O$:

• Nitrogen (N): Left - hand side has $4$ N atoms (from $4NH_3$), right - hand side has $4$ N atoms (from $4NO$).
• Hydrogen (H): Left - hand side has $4\times3 = 12$ H atoms (from $4NH_3$), right - hand side has $6\times2=12$ H atoms (from $6H_2O$).
• Oxygen (O): Left - hand side has $5\times2 = 10$ O atoms (from $5O_2$), right - hand side has $4 + 6=10$ O atoms (from $4NO$ and $6H_2O$). So it obeys the law.

Step2: Check atoms in reaction B

For $8H_2S+8Cl_2
ightarrow S_8 + 16HCl$:

• Hydrogen (H): Left - hand side has $8\times2=16$ H atoms (from $8H_2S$), right - hand side has $16$ H atoms (from $16HCl$).
• Sulfur (S): Left - hand side has $8$ S atoms (from $8H_2S$), right - hand side has $8$ S atoms (from $S_8$).
• Chlorine (Cl): Left - hand side has $8\times2 = 16$ Cl atoms (from $8Cl_2$), right - hand side has $16$ Cl atoms (from $16HCl$). So it obeys the law.

Step3: Check atoms in reaction C

For $CH_4+2O_2
ightarrow CO_2+H_2O$:

• Carbon (C): Left - hand side has $1$ C atom (from $CH_4$), right - hand side has $1$ C atom (from $CO_2$).
• Hydrogen (H): Left - hand side has $4$ H atoms (from $CH_4$), right - hand side has $2$ H atoms (from $H_2O$). This reaction is not balanced, so it does not obey the law.

Step4: Check atoms in reaction D

For $HCl + K_2CO_3
ightarrow KCl+H_2O + CO_2$:

• Hydrogen (H): Left - hand side has $1$ H atom (from $HCl$), right - hand side has $2$ H atoms (from $H_2O$). This reaction is not balanced, so it does not obey the law.

Step5: Check atoms in reaction E

For $2Al(OH)_3+3H_2SO_4
ightarrow Al_2(SO_4)_3+6H_2O$:

• Aluminum (Al): Left - hand side has $2$ Al atoms (from $2Al(OH)_3$), right - hand side has $2$ Al atoms (from $Al_2(SO_4)_3$).
• Hydrogen (H): Left - hand side has $2\times3+3\times2 = 12$ H atoms (from $2Al(OH)_3$ and $3H_2SO_4$), right - hand side has $6\times2 = 12$ H atoms (from $6H_2O$).
• Oxygen (O): Left - hand side has $2\times3+3\times4=18$ O atoms (from $2Al(OH)_3$ and $3H_2SO_4$), right - hand side has $3\times4 + 6=18$ O atoms (from $Al_2(SO_4)_3$ and $6H_2O$).
• Sulfur (S): Left - hand side has $3$ S atoms (from $3H_2SO_4$), right - hand side has $3$ S atoms (from $Al_2(SO_4)_3$). So it obeys the law.

Answer:

A. $4NH_3 + 5O_2
ightarrow4NO + 6H_2O$
B. $8H_2S+8Cl_2
ightarrow S_8 + 16HCl$
E. $2Al(OH)_3+3H_2SO_4
ightarrow Al_2(SO_4)_3+6H_2O$