2nal + cl2 → 2naci + i2

how many grams of iodine are liberated if 546 g of chlorine react with excess sodium iodide?

(i'm a really busy ap student with a lot of homework. just need the work and answer! )

Mass of iodine = 1954.26 g

Explanation:

Given data:

Mass of chlorine react = 546 g

Mass of iodine produced = ?

Solution:

Chemical equation:

2NaI + Cl₂  →  2NaCl + I₂

Number of moles of chlorine:

Number of moles = mass/molar mass

Number of moles = 546 g/71 g/mol

Number of moles = 7.7 mol

Now we will compare the moles of chlorine with iodine.

                     Cl₂          :          I₂

                      1             :          1

                      7.7         :          7.7

Mass of iodine:

Mass = number of moles × molar mass

Mass = 7.7 mol × 253.8 g/mol

Mass = 1954.26 g

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890

the answer is   c.) 1575 j.

explanation: