Question #975d6

1 Answer
Apr 19, 2016

Answer:

You need #"1400 kg NaCl"#.

Explanation:

Molar mass of chlorine#("Cl"_2) = "71 g·mol"^-1#.

Molar volume of chlorine at STP #"= 22.4 L"#

Molar mass of #"NaCl" = "(23.0 + 35.5) g·mol"^-1 = "58.5 g·mol"^-1#

1 mol #"NaCl"# produces #"0.5 mol"# chlorine or #"0.5 mol" xx "22.4 L·mol"^-1 = "11.2 L"# chlorine

So, available #"Cl"_2 # is #"11.2 L/mol of NaCl"# at STP.

So, to have #2.7 xx 10^5color(white)(l) "L Cl"_2 # at STP, we require

#2.7 xx 10^5color(white)(l) "L Cl"_2 xx "1 mol NaCl"/("11.2 L Cl"_2) = 2.4 xx 10^4color(white)(l) "mol NaCl"#

#2.4 xx 10^4color(white)(l) "mol" xx "58.5 g/mol" = 140 xx 10^4color(white)(l) "g NaCl" = "1400 kg NaCl"#