Question

A solution containing 20.0 g of sodium sulfite reacts with 7.0 ml of phosphoric acid. The concentration of the acid solution is such that there are 1.83 grams of H3PO4 per milliliter of solution. Determine the following: a. The mass of the excess reactant remaining at completion. b. Grams of water produced. c. Moles of sodium phosphate produced. d. Grams of sulfur dioxide produced.

Answer 1

a) The mass of the excess reactant remaining at completion is 7.36 grams.

b) mass of water produced is 3.51 grams.

c) moles of sodium phosphate produced is 0.0186.

d) grams of sulphur dioxide produced is 12.49 grams.

Step-by-step explanation:

Data given:

mass of sodium sulphite = 20 grams

volume of phosphoric acid = 7 ml

concentration of the phosphoric acid is = 1.83 grams/ml

mass of excess reactant =?

grams of water produced =?

moles of sodium phosphate =?

grams of sulphur dioxide =?

balance equation for the reaction:

2
`H_(3)`P
`O^(4)` + 3
`Na_(2)`
`SO_(3)` ⇒ 2
`Na_(3) PO_(4)` +3
`SO_(2)` +3
`H_(2)`O

Now the number of moles will be calculated by using the formula:

molarity =
`(number of moles)/(volume of solution)`

For phosphoric acid number of moles (atomic mass of H3PO4 = 97.99 grams/mole), mass is 1 ml has 1.83 grams so 7 ml wil have 12.81 grams

number of moles =
`(mass)/(atomic mass of 1 mole)`

=
`(12.81)/(97.99)`

= 0.13 moles of H3PO4 in 7 ml

number of moles of sodium sulphite (atomic mass = 126.04 grams/mole)

=
`(20)/(126.04)`

= 0.15 moles

from the stoichiometry:

limiting reagent is the one which yield low amount of product.

3 moles of phosphoric acid gave 2 moles of Na3PO4

0.13 moles give x moles

`(2)/(3)` =
`(x)/(0.13)`

3x = 0.26

x = 0.086 moles of Na3PO4

2 moles of phosphoric acid gave 3 moles of sulphur dioxide and 3 moles of water

0.13 moles of phosphoric acid will give x moles of water and sulphur dioxide.

`(3)/(2)` =
`(x)/(0.13)` (SO2) and
`(3)/(2)` =
`(x)/(0.13)`

so 0.195 moles of water and 0.195 moles of SO2 is formed.

Now the second reactant:

3 moles of 3
`Na_(2)`
`SO_(3)` gave 2 moles of Na3PO4

So, 0.15 moles will give

`(2)/(3)` =
`(x)/(0.15)`

x = 0.1 moles of Na3P04

3 moles
`Na_(2)`
`SO_(3)` gives 3 mole of water and SO2

so 0.15 mole of water and SO2 formed

so the limiting reagent is H3P04

mass of the products:

mass = number of moles x atomic mass

b) mass of water = 0.195 x 18

= 3.51 grams

c) mass of SO2 = 0.195 x 64.06

= 12.49 grams

d ) mass of sodium phosphate = 0.086 x 163.94

= 14.09 grams

a) mass of excess reactant

Sodium sulphite is excess reactant,

so we had started with 0.15 sodium sulphite

ratio of excess and limiting reagent as moles of sodium phosphate from sodium sulphite is 0.1 and from H3P04 is 0.086 moles so excess is

0.1 x 126.04

= 12.64 grams

we started with 20 grams and used 12.64 grams

excess reagent = 20 - 12.64

= 7.36 grams is the excess reagent.