Question

How many moles of sodium bicarbonate is needed to neutralize 0.8 ml of sulphuric acid?

Answer 1

Answer:

n NaHCO3 = 9.6 E-3 mol

Explanation:

balanced reaction:

• 2 NaHCO3(s) + H2SO4(ac) ↔ Na2SO4(ac) + 2 CO2(g) + 2 H2O(l)

• assuming a concentration of H2SO4 6M....normally worked in the lab

⇒ n H2SO4 = 8 E-4 L * 6 mol/L = 4.8 E-3 mol H2SO4

according to balanced reaction, we have that for every mol of H2SO4 there are two mol of NaHCO3 ( sodium bicarbonate)

⇒ mol NaHCO3 = 4.8 E-3 mol H2SO4 * ( 2 mol NaHCO3 / mol H2SO4 )

⇒ ,mol NaHCO3 = 9.6 E-3 mol

So 9.6 E-3 mol NaHCO3,  are the minimun moles necessary to neutralize the acid.

Answer 2

The number of moles that is required to neutralize 0.8ml of sulphuric acid is 0.03 moles

What is a neutralization reaction?

Neutralization reaction is a reaction where acid reacts with base to form salt and water only.

The reaction is

$2NaHCO_{3} + H_{2}SO_{4} = Na_{2}SO_{4} + 2H_{2}O + 2CO_{2}$

Density of pure sulphuric acid is 1.8391 g/ml

mass of sulphuric acid that reacted

= volume x density

= 0.8 ml x 1.8931 g/ml

= 1.47 g

molar mass of sulphuric acid = 98 g

Number of moles = $\frac{1.47}{98}$ = 0.015 moles

From the equation of the reaction 2 moles of sodium bicarbonate reacted with 1 mole of sulphuric acid, this gives a ratio 2:1.

Number of moles of sodium bicarbonate required is

2 * 0.05 = 0.03 moles

Read more on neutralization reaction here: https://brainly.com/question/203541