1 how many moles of sodium bicarbonate are needed to neutralize 0.9ml of sulphuric acid at stp

0.0338 moles of sodium bicarbonate are required to react with 0.9 mL 100% sulphuric acid solution

Explanation:

Equation of the reaction:

Na₂CO₃(aq) + H₂SO₄(aq) ---> Na₂SO₄(aq) + H₂O(l) + CO₂(g)

Since the concentration of the sulphuric acid is not given in the question, the assumption is made that the sulphuric acid solution is 100% sulphuric acid solution.

The density of 100% concentrated sulfuric acid is 1.839 g/mL.

Amount of acid in 0.9 mL solution = 0.9 mL * 1.839 g/mL = 1.655 g

Number of moles of acid in 1.655 g = mass/molar mass

Molar mass of  H₂SO₄ = 98 g/mol

Number of moles of acid in 1.471 g = 1.655 g / 98 g/mol = 0.0169 moles

From the equation of reaction,  1 mole of H₂SO₄  reacts with 2 moles of Na₂CO₃

0.0169 moles of H₂SO₄  will react with 0.0169 * 2 moles of Na₂CO₃ = 0.0338 moles

Therefore, 0.0338 moles of sodium bicarbonate are required to react with 0.9 mL 100% sulphuric acid solution

it is a solid

explanation: