An auto mechanic spills 67 mL of 2.6 M H2SO4 solution from a rebuilt auto battery. How many milliliters of 1.3 M NaHCO3 must be poured on the spill to react completely with the sulfuric acid?

2 NaHCO₃ + H₂SO₄  →  Na₂SO₄ + 2 H₂O + 2 CO₂

we are supposed to find the volume of a 1.3 M solution, which has twice the number of moles as H2SO4 in 67 mL of 2.6M solution; since we need 2 moles of NaHCO₃ to react with 1 mole

Number of Moles of H₂SO₄ Spilled:

Since the molarity is 2.6M,

We have 2.6 moles in 1000 mL of the solution

We have 0.26 moles in 100 mL of the solution

Moles in 67mL:

Since we have 0.26 moles in 100 mL

to get the number of moles in 67 mL, we have to find 67% of 0.26

Moles in 67 mL = 67 * 26 / 10000 = 0.1742 moles OR 0.17 moles (approx)

Volume of NaHCO₃ which has (0.17*2) = 0.34 moles:

The molarity of NaHCO₃ is 1.3 M, So:

We have 1.3 moles in 1000 mL

We have 0.13 moles in 100 mL

To get the volume required, we have to find what percent of 0.13 is 0.17:

x*13/100 = 34

13x = 3400

x = 3400 / 13

x = 261.5 %

since 261.5% of 0.13 is 0.34

So, volume which contains 0.34 moles = 261.5% of the volume that contains 0.13 moles

Volume = 261.5% * 100

Volume = 261.5 mL

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.