(05.04 MC) The following reaction shows sodium hydroxide reacting with sulfuric acid.

NaOH + H2SO4 → Na2SO4 + H2O

How many grams of Na2SO4 are produced from 10.0 grams of NaOH?

(Molar mass of Na = 22.989 g/mol, O = 15.999 g/mol, H = 1.008 g/mol, S = 32.065 g/mol) (4 points)

a
17.8 grams

b
19.2 grams

c
35.5 grams

d
38.5 grams

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17.8 g Na2SO4

Explanation:

1. Write the balanced equation

4NaOH + 2H2SO4 -> 2Na2SO4 + 4H2O

2. Begin Stoichiometry using the molar mass of the starting substance

The molar mass is listed on the periodic table as mass.

Na-22.990

O-15.999

H-1.008

They do not have any subscripts so you can just add them up to get the molar mass for the compound

NaOH-39.997

You have to put 1 mole at the top because that's the molar mass of 1 NaOH

3. Create a mole ratio from the the coefficients of the two compounds of interest

You are using NaOH and Na2SO4. NaOH has a coefficent of 4 and Na2SO4 has a coefficent of 2.

Na2SO4 goes on top because that's what we're finding

4. Include it in the stoichiometry conversion

*

5. Include the conversion factor for the molar mass of the ending substance

The conversion factor will have Na2SO4 because that's what we need to find, in grams specifically. So the conversion factor is the molar mass of Na2SO4 over 1 mole

*

6. Solve with dimensional analysis. Cancel out all units except for the final substance

The final substance will be g Na2SO4

(10*1*2*142.036) / (1*39.997*1*1)

17.75 g Na2SO4

17.8 g Na2SO4

Hope this helps! If you have further questions, don't be afraid to reach out for help. Have a nice day.