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Andrej [43]
3 years ago
10

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

Chemistry
2 answers:
viva [34]3 years ago
5 0
<span>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.
</span><span>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.</span>
muminat3 years ago
5 0

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.

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Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of NH_3 = 100 g

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