Respuesta :
CS2(l)+Cl2(g)=CCl4(l)+S2Cl2(l)
1 C atom gets in , 1 C atom gets out
2 S atoms get in , 2 S atoms get out
2Cl atoms get in , 6 Cl atoms get out => we have to put a 3 in front of Cl2
Balanced equation:
CS2(l)+3Cl2(g)=CCl4(l)+S2Cl2(l)
1 C atom gets in , 1 C atom gets out
2 S atoms get in , 2 S atoms get out
2Cl atoms get in , 6 Cl atoms get out => we have to put a 3 in front of Cl2
Balanced equation:
CS2(l)+3Cl2(g)=CCl4(l)+S2Cl2(l)
Answer:
CS₂(l) + 3Cl₂(g) → CCl₄(l) + S₂Cl₂(l)
Explanation:
The next logical step is to list the number of C, Cl, and S atoms on the reactant and product side of the equation.
CS₂(l) + Cl₂(g) → CCl₄(l) + S₂Cl₂(l)
There is 1 C atom on both the reactant and product side of the equation.
There are two 2 S atoms on both the reactant and product side of the equation.
There are 2 Cl atoms on the reactant side and 6 (4+2) Cl atoms on the product side. Therefore to balance the equation we have to place a 3 coefficient before Cl₂ on the reactant side of the equation.
Therefore the final balanced equation is,
CS₂(l) + 3Cl₂(g) → CCl₄(l) + S₂Cl₂(l)