1)Aluminum reacts with bromine to produce aluminum bromide. How any moles of bro

Question

1)Aluminum reacts with bromine to produce aluminum bromide. How any moles of bromine are required to react with 3.53 moles of aluminum? Record your answer to two (2) decimal places.

2)Consider a 16.00 g sample of lithium carbonate (Li2CO3). How many grams of oxygen are present in the sample? Record your answer to three (3) decimal places.

3)What is the molarity of NaCl in a solution made by adding 27.13 g of NaCl to enough water to make 435.9 mL of solution? Record your answer to three (3) decimal places.

Explanation / Answer

2Al + 3Br2 <--> 2AlBr3

3 mol of Br2 are needed to form 2 mol of Al

3/2*3.53 = 5.295 mol of Br are needed

5.30 mol of Br

2)

m = 16g of Li2CO3

grams of O...

MW of Li2CO3 = 73.9

MW of O = 16

Mol O = 48

48/73.9 = 0.649

we have 16 g then 16*0.649 = 10.384 grams

3)

M = mol / volume

m = 27.13 NaCl

MW = 58.4 g

mol NaCL = 27.13/58.4 = 0.4645 mol

V = 435.9 or 0.4359 L

M = 0.4645 / 0.4359 = 1.0656 M

M = 1.066 M

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