You have seven closed containers, each with equal masses of chlorine gas (cl2). you add 10.0 g of sodium to the first sample, 20.0 g of sodium to the second sample, and so on (adding 70.0g of sodium to the seventh sample). sodium and chlorine react to form sodium chloride according to the equation: 2na(s) + cl2(g)  2nacl(s) after the reaction is complete, you collect and measure the amount of sodium chloride

The reaction between Na and Cl2 is

2Na + Cl2 ---> 2NaCl

So two moles of Na will react with one mole of Cl2

However here in the given question the moles of Cl2 is constant as mass is constant. Say we have 71 of Cl2 in each container, the moles = 71 / 35.5 = 2 moles of Cl2

Moles of Na = Mass / atomic mass

The container will have following moles of Na

1 ) Moles = 10 / 23 = 0.435  (these will react with 0.2175 moles of Cl2 to give 0.435 moles of NaCl]

2) moles = 20 / 23 = 0.87 ( (these will react with 0.435 moles of Cl2 to give 0.87 moles of NaCl]

3) moles = 30 / 23 = 1.3 ( (these will react with 0.65 moles of Cl2 to give 1.3 moles of NaCl]

4) moles = 40 / 23 = 1.74 (these will react with 0.87 moles of Cl2 to give 1.7 moles of NaCl]

5) moles = 50 / 23 = 2.18 (these will react with 1.09 moles of Cl2 to give 2.18 moles of NaCl]

6) moles = 60 / 23 = 2.61  (these will react with 1.3 moles of Cl2 to give 2.61 moles of NaCl]

7) moles = 70 / 23 =  3 moles  (these will react with 1.5 moles of Cl2 to give 3 moles of NaCl]

i think   the answer is c hope i'm right

the answer is c. nacl