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Ksju [112]
3 years ago
14

How much excess reactant remains from the reaction of 8.3 g of sodium and 4.5 g of chlorine?

Chemistry
1 answer:
Orlov [11]3 years ago
5 0
Sodium:
moles = mass/RFM = 8.3/23 = 0.3608..

Chlorine:
moles = mass/RFM = 4.5/71 = 0.06

Ratio of Na to Cl is 2:1
So 0.06 x 2 = 0.12

So we know that the Na is in excess

Now we have to figure out by how much:

Mass = moles x RFM = 0.12 x RFM of 2NaCl = 0.12 x 58.5 = 7.4 (roughly)

Now to figure out by how much do 8.3 + 4.5 = 12.7 and do 12.7 - 7.4 = 5.3 (roughly)

So your answer is A
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