Answer:
10.3 grams
Step-by-step explanation:
The balanced equation shows that 1 mole of Cl2 reacts with 2 moles of NaBr. To find out how much Cl2 is required to react with 30 grams of NaBr, we need to convert grams to moles.
First, we need to find the molar mass of NaBr:
NaBr = 23 + 79.9 = 102.9 g/mol
Now we can calculate the number of moles of NaBr:
30 g NaBr ÷ 102.9 g/mol = 0.291 moles NaBr
From the balanced equation, we know that 1 mole of Cl2 reacts with 2 moles of NaBr. Therefore, we need half as many moles of Cl2 as we have moles of NaBr:
0.291 moles NaBr ÷ 2 = 0.1455 moles of Cl2
Finally, we can convert moles of Cl2 to grams using its molar mass:
Cl2 = 35.5 x 2 = 71 g/mol
0.1455 moles Cl2 x 71 g/mol = 10.3 grams of Cl2
Therefore, 10.3 grams of Cl2 are required to react completely with 30 grams of NaBr in this reaction.