Answer:
2,849,370
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:
2.85 g
Explanation:
To solve this problem, we need to first write out a balanced equation for the reaction:
[tex]\boxed{\mathrm{Ca(OH)_2 + H_2SO_4 \rightarrow CaSO_4 + 2H_2O}}[/tex]
Next, we have to calculate the number of moles of [tex]\mathrm{H_2SO_4}[/tex] that will be neutralized:
no. of moles of [tex]\mathrm{H_2SO_4}[/tex] = concentration × volume/1000
= 0.850 × [tex]\frac{45.3}{1000}[/tex]
= 0.0385 mol
As we can see from the balanced equation above, the molar ratio of [tex]\mathrm{Ca(OH)_2}[/tex] and [tex]\mathrm{H_2SO_4}[/tex] are equal; therefore their mole numbers are also equal.
This means that 0.0385 moles of [tex]\mathrm{Ca(OH)_2}[/tex] will be required to neutralize the [tex]\mathrm{H_2SO_4}[/tex].
Now we can calculate the mass of [tex]\mathrm{Ca(OH)_2}[/tex] 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 [tex]\mathrm{Ca(OH)_2}[/tex] will be needed.
