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

8.0356 * 10^-5 moles of NaHCO3

Explanation:

Sulphuric acid = H2SO4

Sodium bicarbonate = NaHCO3

The reaction between both compounds is given by;

2NaHCO3(aq) + H2SO4(aq) → Na2SO4(aq) + 2CO2(g) + 2H2O(l)

In the reactin above;

2 mol of NaHCO3 neutralizes 1 mol of H2SO4

At stp, 1 mol occupies 22.4 L;

1 mol = 22.4 L = 22400 mL

x mol = 0.9 mL

x = 0.9 / 22400 = 4.0178 * 10^-5 moles of H2SO4

Since 2 mol = 1 mol from the equation;

x mol = 4.0178 * 10^-5

x mol = 2 * 4.0178 * 10^-5

x = 8.0356 * 10^-5 moles of NaHCO3

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.