Good evening everyone, need some help completing this lab. INFO Mass of \"contai

ID: 881804 • Letter: G

Question

Good evening everyone, need some help completing this lab.

INFO Mass of "container" 104.707g

Mass of "container" + Na2CO3 106.505g

Mass of "container" + NaCL 1st heating 106.625

Mass of "container" + NaCL 2nd heating 105.918

1- From your data, alculate the mols of Na2CO3 and mols of NaCL. Show all calculations work.

2. Calculate teh simplest whole number mole ratio of Na2CO3: NaCL. Show all calculation.

3. How does yoru experimental ratio compare to the accepted 1:2 ratio? If they are different, explain what might have caused the difference.

4. Calculate teh percent error. %E = accepted value - exp erimental value / accepted value X 100%

Please please help and TIA

Lets calculate mass of Na2CO3

Mass of Na2CO3 = Mass of container + Na2CO3 - Mass of container

=(106.505-104.707 )g= 1.798 g

Mass of NaCl = (Mass of NaCl + container ) – Mass of container

= 105.918 g – 104.707 g

= 1.211 g NaCl

Lets calculate moles:

Mol Na2CO3 = 1.798 g * 1mol Na2CO3 / Molar mass of Na2CO3

=1.798 g * 1mol Na2CO3/ 105.9888 g per mol

=0.01696 mol

Mol NaCl =1.211 g NaCl * 1mol NaCl / 58.44 g

= 0.02072 mol NaCl

To find simple whole number ratio we divide mol NaCl and Na2CO3 by mol of Na2CO3

Mol Na2CO3= 0.01696 mol / 0.01696 mol = 1 mol Na2CO3

Mol NaCl =0.02072 mol / 0.01696 mol = 1.22

This simply show the ratio would be 1 : 1

This is different from 1:2

The reason :

No proper heating of Na2CO3 , or loss of NaCl

Percent error = l (1/2) – (1/1) / (1/2 )* 100 l

= 10.0 %

Percent error would be 10.0 %

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