Question

Find the mass of 50% CaCO3 which will produce 136 g of CaSO4? ​

Answer 1

Answer:

68g bro

Explanation:

You can also do by unitary method

Answer 2

The mass of 50% $CaCO_3$ is 200 g

Explanation:

The given equation is

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

From this it is clear that 1 mole of $CaCO_3$ will produce 1 mole of $CaSO_4$

Given that mass of $CaSO_4$ produced = 136 g

Molar weight of $CaSO_4$ = 136

Thus, moles of $CaSO_4$ produced = Mass/Molar mass = 136/136 = 1 mole

Thus, moles of $CaCO_3$ utilised = 1

We know that in 1 mole there are 100 gram of $CaCO_3$

Thus, mass of $CaCO_3$ = 100

But it is given that $CaCO_3$ is only 50% pure

Thus, total mass of $CaCO_3$ = 200 g

Hope this answer is helpful.

Know More:

Q: What volume of carbon dioxide at NTP will be produced by the reaction of 2 litres of HCl solution (sp. gravity = 1.16 and strength =30% by weight) on excess calcium carbonate? (full process) needed​

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