Question

3. You place 2.80 g of phosphoric acid into a 25.0 mL of a 1.25M sodium hydroxide solution. The molar mass of phosphoric acid =98.00 g/mole, sodium hydroxide =40.01 g/mole and water is 18.02 g/mole. Answer the following questions. (6 points) H 3

​
PO 4
​
+3NaOH→3H 2
​
O+Na 3
​
PO 4
​
a. Determine the mass of water that is produced. H 3
​
PO 4
​
+3NaOH→3H 2
​
O+Na 3
​
PO 4
​
b. Determine the mass of the reactant that remains after the chemical reaction is complete.

Answer 1

Step-by-step explanation:

a)

from the formula you know that

1 mol h3po4 + 3 mol naoh gives 3 moles h2o

you have

`(2.80g)/(98 (g)/(mol) ) = 0.0286mol \: h3po4 \\ 1.25 * (25ml)/(1000 (ml)/(l) ) = 0.03125 \: mol \: naoh`

how many moles of h2o will you produce with the reactants you have?

`(1 \: mol \: h3po4)/(3 \: mol \: h2o) = (0.0286 \: mol \: h3po4)/(x \: mol \: h2o)`

x mol h2o = 0.0858 mol

`(3 \: mol \: naoh)/(3 \: mol \: h2o) = (0.03125 \: mol \: naoh)/(x \: mol \: h2o)`

x mol h2o = 0.03125 mol

one of your reactants gives less moles of h2o , that's your limiting reactant and that means you'll get 0.03125 mol of h2o and not 0.0858 mol

so you'll have

0.03125 mol × 18 g/mol h20 = 0.5625 g h2o

b)

the mass of reactant that remains is your reactant in excess, h3po4

since you'll only produce 0.03125 mol h2o

you'll only consume x moles of h3po4

`(1 \: mol \: h3po4)/(3 \: mol \: h2o) = (x \: mol \: h3po4)/(0.03125 \: mol \: h2o)`

0.0104 mol = x mol h3po4

you'll have an excess of h3po4 that won't react

moles used - moles consumed = moles in excess

0.0286 mol - 0.0104 mol = 0.0182 mol

the mass of the reactant that remains is

0.0182 mol × 98g/mol = 1.78g h3po4