Question

A sample contains 45% CaCO3, 40% MgCO3 and rest is clay. 10g of this sample treated with H2SO4. Calculate min. mass of H2SO4 of 15% by mass required to decompose the sample completely. Answer fast..... Don't take much time to answer any of the questions...........

Answer 1

Answer: The mass of $H_2SO_4$ required to decompose the given amount of sample is 60 grams.

Explanation:

We are given:

Mass of the sample = 10 g

Mass percent of $CaCO_3=45\%=\frac{45}{100}\times 10=4.5g$

Mass percent of $MgCO_3=40\%=\frac{40}{100}\times 10=4g$

Mass of clay = $10-(4.5+4)=1.5g$

As, we know clay does not react with sulfuric acid, so the chemical equation for the reaction of calcium carbonate and magnesium carbonate with sulfuric acid follows:

$MgCO_3+CaCO_3+2H_2SO_4\rightarrow MgSO_4+CaSO_4+2H_2O+CO_2$

By Stoichiometry of the reaction:

1 mole of $CaCO_3$ that is 100 g reacts with 1 mole of $MgCO_3$ i.e. 84 g and 2 moles of $H_2SO_4$ i.e. 2(98) = 196g

In total, 184 g of $(CaCO_3+MgCO_3)$ reacts with 196 g of sulfuric acid.

So, 8.5 g of $(CaCO_3+MgCO_3)$ will react with $\frac{196}{184}\times 8.5=9g$ of sulfuric acid.

To calculate the amount of sulfuric acid by 15 % of sulfuric acid, we use the equation:

$100\times 9=15\times M_{H_2SO_4}\\\\M_{H_2SO_4}=60g$

Hence, the mass of $H_2SO_4$ required to decompose the given amount of sample is 60 grams.

Answer 2

Answer:

Hope this helps you...........................