In the reaction #2NaOH+H_2SO_4 -> 2H_2O+Na_2SO_4#, how many grams of sodium sulfate will be formed if you start with 200.0 grams of sodium hydroxide and you have an excess of sulfuric acid?

2 Answers
Mar 15, 2016

Answer:

Use the mole analogy of the reactant with the product. and substitute with the mole definition.

Answer is 355 grams.

Explanation:

Given the molecular weights:

#Mr_(NaOH)=40 g/(mol)#
#Mr_(Na_2SO_4)=142 g/(mol)#

The analogy of the moles will be held constant:

#n_(NaOH)/n_(Na_2SO_4)=2/1#

#n_(NaOH)/n_(Na_2SO_4)=2#

For each one, substitute:

#n=m/(Mr)#

Therefore:

#n_(NaOH)/n_(Na_2SO_4)=2#

#(m_(NaOH)/(Mr_(NaOH)))/(m_(Na_2SO_4)/(Mr_(Na_2SO_4)))=2#

#(200/40)/(x/142)=2#

#(200*142)/(40x)=2#

#200*142=2*40x#

#x=(200*142)/(2*40)=(100*142)/40=10*142/4=1420/4=#

#=710/2=355grams# (or just use a calculator)

Mar 15, 2016

Answer:

Through conversion method using mole ratio and formula masses the answer is #355# g of #Na_2SO_4#

Explanation:

  1. Equation is already balanced, therefore what you need to find are the formula masses of the involved compounds, #NaOH# and #Na_2SO_4#;
  2. Once known, start the calculation by converting #200# g #NaOH# to mole #NaOH# by multiplying it with the ratio of the formula mass of #NaOH#;
  3. The result from the above calculation, will then be multiplied by the mole ratio of #Na_2SO_4# and #NaOH#, which is #(1 mol Na_2SO_4)/(2 mol NaOH)#;
  4. Since, we are asked to find the mass of #Na_2SO_4# formed in this reaction, we need to multiply the answer of #step 3# to the ratio of the formula mass of #Na_2SO_4#.
  5. Per calculation, the answer in mass is #355# grams of #Na_2SO_4#.