Question

compute the mass of caso4 that can be prepared by the reaction of 3.2900g of h2so4 with 3.1660g of caco3

Answer 1

1 mole of calcium carbonate reacts with 1 mole of sulfuric acid and produces 1 mole of calcium sulfate. 3.1660 g of CaCO3 is how many moles of calcium carbonate? 3.1660 / 100.0869 = 0.031633 moles. 3.2900 g of H2S04 is how many moles of sulfuric acid? 3.2900 / 98.079 = 0.033544 moles.
The lesser of the two is 0.031633 moles. Therefore, 0.031633 moles of calcium carbonate will combine with 0.031633 moles of sulfuric acid to produce 0.031633 moles of calcium sulfate. Molecular weight of calcium sulfate is 136.14 g/mol. Therefore, 0.031633 moles of calcium sulfate will weight 0.031633 x 136.14 g/mol = 4.3065 grams.

Answer 2

Answer : The mass of $CaSO_4$ produced will be, 4.302 grams.

Explanation : Given,

Mass of $H_2SO_4$ = 3.2900 g

Mass of $CaCO_3$ = 3.1660 g

Molar mass of $H_2SO_4$ = 98.079 g/mole

Molar mass of $CaCO_3$ = 100.087 g/mole

Molar mass of $CaSO_4$ = 136.14 g/mole

First we have to calculate the moles of $H_2SO_4$ and $O_2$.

$\text{Moles of }H_2SO_4=\frac{\text{Mass of }H_2SO_4}{\text{Molar mass of }H_2SO_4}=\frac{3.2900g}{98.079g/mole}=0.0335moles$

$\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}=\frac{3.1660g}{100.087g/mole}=0.0316moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$H_2SO_4+CaCO_3\rightarrow CaSO_4+H_2CO_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCO_3$ react with 1 mole of $H_2SO_4$

So, 0.0316 moles of $CaCO_3$ react with 0.0316 moles of $H_2SO_4$

From this we conclude that, $H_2SO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCO_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CaSO_4$.

As, 1 mole of $CaCO_3$ react to give 1 moles of $CaSO_4$

So, 0.0316 moles of $CaCO_3$ react to give 0.0316 moles of $CaSO_4$

Now we have to calculate the mass of $CaSO_4$.

$\text{Mass of }CaSO_4=\text{Moles of }CaSO_4\times \text{Molar mass of }CaSO_4$

$\text{Mass of }CaSO_4=(0.0316mole)\times (136.14g/mole)=4.302g$

Therefore, the mass of $CaSO_4$ produced will be, 4.302 grams.