Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resulting solution. The sodium bicarbonate reacts with sulfuric acid according to: \begin{equation} \begin{array}{r}{2 \mathrm{NaHCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+} \quad\\\ {2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g)}\end{array} \end{equation} Sodium bicarbonate is added until the fizzing due to the formation of \(\mathrm{CO}_{2}(g)\) stops. If 27 \(\mathrm{mL}\) of 6.0 \(\mathrm{MH}_{2} \mathrm{SO}_{4}\) was spilled, what is the minimum mass of \(\mathrm{NaHCO}_{3}\) that must be added to the spill to neutralize the acid?

Step by step solution

01

Find the moles of H2SO4

First, we need to calculate the moles of H2SO4 using the given volume (27 mL) and molarity (6.0 M). The formula to calculate moles is: Moles = Molarity × Volume (in Liters) So, moles of H2SO4 = 6.0 M × (27 mL × 0.001 L/mL) = 6.0 M × 0.027 L = 0.162 moles of H2SO4

02

Find the moles of NaHCO3 needed

Now that we have the moles of H2SO4, we'll use the balanced chemical equation to find the corresponding moles of NaHCO3 needed for neutralization: 2 NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l) + 2 CO2(g) From the balanced equation, it's clear that 2 moles of NaHCO3 are required to neutralize 1 mole of H2SO4. So, the moles of NaHCO3 required = 2 × Moles of H2SO4 = 2 × 0.162 = 0.324 moles of NaHCO3

03

Convert moles of NaHCO3 to mass

Finally, we'll convert the moles of NaHCO3 to mass using the molar mass of NaHCO3. We can find the molar mass by adding the atomic masses of each element in the compound: Molar mass of NaHCO3 = (22.99 g/mol Na) + (1.008 g/mol H) + (12.01 g/mol C) + (3 × 16.00 g/mol O) = 84.007 g/mol Now, we can convert moles of NaHCO3 to mass: Mass of NaHCO3 = Moles × Molar mass = 0.324 moles × 84.007 g/mol = 27.22 g (approximately) So, the minimum mass of NaHCO3 needed to neutralize the spill is approximately 27.22 g.

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