consider the following intermediate chemical equations.\n2p(s) + 3cl₂(g) → 2pcl₃(l)\npcl₃(l) + cl₂(g) →…

consider the following intermediate chemical equations.\n2p(s) + 3cl₂(g) → 2pcl₃(l)\npcl₃(l) + cl₂(g) → pcl₅(s)\nwhen you combine the intermediate chemical equations, which substance do you cancel out?

consider the following intermediate chemical equations.\n2p(s) + 3cl₂(g) → 2pcl₃(l)\npcl₃(l) + cl₂(g) → pcl₅(s)\nwhen you combine the intermediate chemical equations, which substance do you cancel out?

Answer

Explanation:

Step1: Write the two given equations

$2P(s)+3Cl_2(g)\rightarrow 2PCl_3(l)$ $PCl_3(l)+Cl_2(g)\rightarrow PCl_5(s)$

Step2: Multiply the second - equation by 2 to balance

$2PCl_3(l)+2Cl_2(g)\rightarrow 2PCl_5(s)$

Step3: Add the first equation and the multiplied second equation

$(2P(s)+3Cl_2(g)\rightarrow 2PCl_3(l))+(2PCl_3(l)+2Cl_2(g)\rightarrow 2PCl_5(s))$ $2P(s)+(3 + 2)Cl_2(g)\rightarrow 2PCl_5(s)$ $2P(s)+5Cl_2(g)\rightarrow 2PCl_5(s)$

Step4: Identify the substance that cancels out

In the addition of the two equations, $PCl_3(l)$ appears on the product side of the first equation and the reactant side of the second - equation. So $PCl_3$ cancels out.

Answer:

$PCl_3$