Iodine is prepared both in the laboratory and commercially by adding cl2(g) to an aqueous solution containing sodium iodide: how many grams of sodium iodide, nai, must be used to produce 48.7 g of iodine, i2?

The balanced equation for the reaction between NaI and Cl₂
2NaI and Cl₂ --> 2NaCl + I₂
stoichiometry of NaI to I₂ is 2:1
this means that when 2 mol of NaI reacts, it produces  1 mol of I₂
Molar mass of I₂ is 254 g/mol 
Number of I₂ moles produced - 48.7 g / 254 g/mol = 0.192 mol
for 1 mol of I₂ to be formed - 2 mol of NaI required
Therefore to form 0.192 mol of I₂ - 2 x 0.192 = 0.384 mol of NaI
Then mass of NaI required - 0.384 mol x 150 g/mol = 57.6 g

Gas is a type of carbon, i think lol