Question

HELP!

How many grams of Ca(OH)2 are needed to neutralize 45.3 mL of 0.850 M H2SO4?

Report your answer with THREE significant figures.

Answer 1

2,849,370

Step-by-step explanation:

1. We'll use the law: [(Molar×Volume)÷ number of moles] of the base, and [(Molar×Volume)÷number of moles] of acids.

2. The answer should be a "mass" (how many grams) of the base, so we'll use for the base this law: (mass÷molar mass×number of moles).

3. The molecular mass (or molar) of Ca(OH)

2 =1(Atomic mass of Ca)+2(Atomic mass of O) +2(Atomic mass of H)

=40+2(16)+2(1)

=40+32+2

=74

4. No. of moles would be 1 in the acid and base due to: 1 Ca(oH)² + 1 H²SO⁴ --> 1 CaSO⁴ +2H²O.

5. Volume should be used in Leter so: 45.3×10³.

6. Put all the given in the question and solve: (mass (x) unknown) ÷ (74×1) = (0.850×45.3×10³) ÷ (1)

7. Then: (x)÷74 = 38,505 --> 2,849,370.

Answer 2

2.85 g

Step-by-step explanation:

To solve this problem, we need to first write out a balanced equation for the reaction:

`\boxed{\mathrm{Ca(OH)_2 + H_2SO_4 \rightarrow CaSO_4 + 2H_2O}}`

Next, we have to calculate the number of moles of
`\mathrm{H_2SO_4}` that will be neutralized:

no. of moles of
`\mathrm{H_2SO_4}` = concentration × volume/1000

= 0.850 ×
`(45.3)/(1000)`

= 0.0385 mol

As we can see from the balanced equation above, the molar ratio of
`\mathrm{Ca(OH)_2}` and
`\mathrm{H_2SO_4}` are equal; therefore their mole numbers are also equal.

This means that 0.0385 moles of
`\mathrm{Ca(OH)_2}` will be required to neutralize the
`\mathrm{H_2SO_4}`.

Now we can calculate the mass of
`\mathrm{Ca(OH)_2}` required:

mass = no. of moles × molar mass

= 0.0385 × [40 + 2×(16+1)]

= 0.0385 × 74

= 2.85 g (3 s.f.)

Therefore, 2.85 g of
`\mathrm{Ca(OH)_2}` will be needed.