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4 years ago

If a particular ore contains 55.1 % calcium phosphate, what minimum mass of the ore must be processed to

Chemistry

1 answer:

Diano4ka-milaya [45]4 years ago

3 0

Answer:

6049.69 g or 6.04969 Kg of ore must be processed to get 1 Kg of phosphorous.

Explanation:

Given

An ore has 51% Calcium phosphate

To find

The minimum mass of ore to be processed to get 1.00 Kg of phosphorous

First find the mass of phosphorous in 1 mole = molar mass of Calcium phosphate, Ca₃(PO₄)₃.

Molar mass of Ca₃(PO₄)₃ is 310 g

Molar mass of P is 31 g

1 mole of Ca₃(PO₄)₃ has 3 atoms of phosphorous

i.e 310 g of Ca₃(PO₄)₃ has 3× 31g of P

= 93 g P

93 g P wil be present in 310 g of Ca₃(PO₄)₃

1 Kg or 1000 g of P will be in (1000÷93) ×310

=3333.33 g of Ca₃(PO₄)₃

but the ore has only 55.1% Ca₃(PO₄)₃

i.e

100 g of Ca₃(PO₄)₃ will have 55.1g Ca₃(PO₄)₃

we need 3333.33g of Ca₃(PO₄)₃

100 g of ore will have 55.1g Ca₃(PO₄)₃

3333.33g of Ca₃(PO₄)₃ will be present in

(3333.33÷ 55.1) ×100

= 6049.60 g of the ore

So 6049.69 g or 6.04969 Kg of ore must be processed to get 1 Kg of phosphorous.

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Answer:

$\boxed{6.08 \times 10^{23}}$

Explanation:

Data:

I = 2.15 A

t = 8 min 24 s

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p = 774.2 To

1. Write the equation for the half-reaction

2H₂O ⟶ O₂ + 4H⁺ + 4e⁻

2. Calculate the moles of oxygen

$p = \text{774.2 To} \times \dfrac{\text{1 atm}}{\text{760 To}} = \text{1.0187 atm}$

V = 0.0654 L

T = (26.0 + 273.15) K = 299.15 K

$\begin{array}{rcl}pV& = & nRT\\1.0189 \times 0.0654 & = & n \times 0.08206 \times 299.15\\0.06662 & = & 24.55n\\\\n & = & \dfrac{0.06662}{24.55}\\\\n & = & 2.714 \times 10^{-3}\\\end{array}$

3. Calculate the moles of electrons

$\text{n} = \text{2.714 $\times 10^{-3}$ mol oxygen} \times \dfrac{\text{4 mol electrons}}{\text{ 1 mol ozygen}} = \text{ 0.01086 mol electrons}$

4. Calculate the number of coulombs

t = 8 min 24 s =504 s

Q = It = 504 s × 2.10 C·s⁻¹= 1058 C

5. Calculate the number of electrons

$\text{No. of electrons} = \text{1058 C} \times \dfrac{\text{1 electron}}{1.602 \times 10^{-19}\text{ C}} = 6.607 \times 10^{21}\text{ electrons}$

6. Calculate Avogadro's number

$N_{\text{A}} = \dfrac{6.607 \times 10^{21}}{0.01086} = \mathbf{6.08 \times 10^{23}}\\\\\text{The experimental value of Avogadro's number is } \boxed{\mathbf{6.08 \times 10^{23}}}$