Question

the number of grams of NAOH that completely neutralises 9.8g of phosphoric acid

Answer 1

Answer: 12 grams of NAOH will completely neutralises 9.8g of phosphoric acid.

A Neutralization reaction occurs when a acid reacts with a base to produce water and a salt (ionic compound). In our case, Phosphoric acid($H_{3}PO_{4}$) will react with sodium hydroxide(Base) to form water and sodium phosphate.

$H_{3}PO_{4} + 3NaOH\rightarrow Na_{3}PO_{4} + 3H_{2}O$

Given grams of $H_{3}PO_{4}$ = 9.8 g

We need to find grams of NaOH.

To find the grams of NaOH we need to follow the following steps.

Step 1 : Convert 9.8 g $H_{3}PO_{4}$ to moles of $H_{3}PO_{4}$

$Moles = \frac{Grams}{Molar mass}$

Molar mass of $H_{3}PO_{4}$ = 98g/mol

$Moles = \frac{9.8 g}{98\frac{g}{mol}}$

Moles of $H_{3}PO_{4}$ = 0.1 mol

Step 2 : Calculate the moles of NaOH using moles of $H_{3}PO_{4}$ and mole ratio.

The mole ratio is the ratio of moles of one substance to the moles of another substance in a balanced equation. Mole ratio are the coefficient(number present in front) of the compound present in the balanced chemical equation.

Coefficient of NaOH is 3 and coefficient of $H_{3}PO_{4}$ is 1 , so mole ratio of NaOH to $H_{3}PO_{4}$ is 3:1

$(0.1 mol H_{3}PO_{4})\times \frac{(3 mol NaOH)}{(1 mol H_{3}PO_{4})}$

$(0.1 )\times \frac{(3 mol NaOH)}{(1)}$

Moles of NaOH = 0.3 mol

Step 3 : convert moles of NaOH to grams of NaOH

Grams = Moles X Molar mass

Molar mass of NaOH = 40 g/mol and Moles of NaOH = 0.3 mol

$Grams = 40\frac{g}{mol}\times 0.3mol$

$Grams = 40g\times 0.3$

Grams of NaOH = 12 grams

Note : The above three steps can also be done in a single step setup.

$(9.8 g H_{3}PO_{4})\times \frac{(1 molH_{3}PO_{4})}{(98 g H_{3}PO_{4})}\times \frac{(3 mol NaOH)}{(1 mol H_{3}PO_{4})}\times \frac{(40 g NaOH)}{(1 mol NaOH)}$

$(9.8)\times \frac{(1)}{(98)}\times \frac{(3)}{(1)}\times \frac{(40 g NaOH)}{(1)}$

Grams of NaOH = 12 grams(g)

Answer 2

Answer:12

Explanation:

3NaOH+H3PO4 gives Na3PO4+3H2O

Molecular mass of 3 NaOH=3*40=120

Molecular mass of 1 Na3PO4=98

120gm NaOH neutralise 98gm Na3PO4

Xgms neutralise 9.8gm

So 9.8*120÷98=12