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koban [17]
3 years ago
7

If 35 ml of 6.0 m h2so4 was spilled, calculate the minimum mass of nahco3 that must be added to the spill to neutralize the acid

.
Chemistry
1 answer:
belka [17]3 years ago
6 0

The balanced equation between the H_2SO_4 and NaHCO_3 is:

H_2SO_4+2NaHCO_3\rightarrow Na_2SO_4+2CO_2+2H_2O

Formula of molarity is:

Molarity = \frac{Moles of solute}{Volume of solution in Liters}

Molarity = 6.0 M, Volume = 35 mL = 0.035 L

Substituting the values,

6 = \frac{Moles of solute}{0.035}

Moles of solute = 0.035 L\times 6 mol/L = 0.21 mole

So, number of moles of H_2SO_4 is 0.21 mole.

From the balanced equation it is clear that for 1 mole of H_2SO_4, 2 moles of NaHCO_3 are required.

Hence, 0.21 mole of H_2SO_4  = 2\times 0.21 mole = 0.42 mole of NaHCO_3

Molar mass of NaHCO_3 = 84.007 g/mol

So, the mass of NaHCO_3 = 84.007 g/mol \times 0.42 mol = 35.283 g




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