Creating Superphosphate

Chemistry Level 2

To create single superphosphate, people use an ore containing $73\% Ca_3(PO_4)_2, 26\% CaCO_3$ and $1\% SiO_2$

People completely dissolve $100kg$ of that ore by using a 65% $H_2SO_4$ solution to create single superphosphate.

How much solution is needed?

Type your answer in kilograms ( $kg$ ). Round your answer to 2 decimal places.

Calculate using the given molar mass of these substances:

$Ca=40, P=31, O=16, C=12, Si=28, H=1, S=32$

The answer is 110.06.

1 solution

Tin Le
Oct 29, 2020

We can calculate that $100kg$ of the ore contains $235 mol$ $Ca_3(PO_4)_2$ and $260 mol$ $CaCO_3$ ( $SiO_2$ is irrevelant because it can't react with $H_2SO_4$ )

Note that single superphosphate must contain $Ca(H_2PO_4)_2$ .

Chemical reactions:

$Ca_3(PO_4)_2 + 2H_2SO_4 \rightarrow Ca(H_2PO_4)_2 + 2CaSO_4 \downarrow$

235 (mol) ------ 470 (mol)

$CaCO_3 + H_2SO_4 \rightarrow CaSO_4 + H_2O + CO_2 \uparrow$

260 (mol) --- 260 (mol)

Deducing from the two chemical reactions, we have to use $m_{H_2SO_4} = n_{H_2SO_4}. M_{H_2SO_4} = (470 + 260).98 = 71540 (g)=71.54 kg$ $H_2SO_4$ .

Therefore, the weight of the needed 65% $H_2SO_4$ solution is: $m=\frac{m_{H_2SO_4}}{65\%} = \frac{71.54}{65\%} \approx \boxed{110.06} (kg)$

Same here. But your thread have some errors (học Văn như thằng này! :) ).

Anh Khoa Nguyễn Ngọc - 7 months, 2 weeks ago

hmu hmu lên lớp làm chmúa hmề giờ dịch In lịt như bị khùng á =))))

Tin Le - 7 months, 2 weeks ago

Creating Superphosphate

The answer is 110.06.

1 solution

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