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3 years ago

Calculate the no. of each atoms presents in 10 grams of calcium carbonate.

Chemistry

1 answer:

raketka [301]3 years ago

8 0

Molar Mass of Calcium carbonate:-

$\\ \sf\longmapsto CaCO_3$

$\\ \sf\longmapsto 40u+12u+3(16u)$

$\\ \sf\longmapsto 52u+48u$

$\\ \sf\longmapsto 100u$

$\\ \sf\longmapsto 100g/mol$

Given Mass=10g

$\boxed{\sf No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}}$

$\\ \sf\longmapsto No\:of\;moles=\dfrac{10}{100}$

$\\ \sf\longmapsto No\:of\:moles=0.1mol$

Now

$\boxed{\sf No\:of\:molecules=No\;of\:moles\times Avagadro\: No}$

$\\ \sf\longmapsto 0.1\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 6.023\times 10^{22}molecules$

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3 years ago

how many grams of calcium sulfate are produced from 10 grams of calcium nitrate and how many grams of calcium sulfate are produc

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Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

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$\text{Moles of} Li_2SO_4=\frac{10g}{110g/mol}=0.091moles$

$Ca(NO_3)_2+Li_2SO_4\rightarrow 2LiNO_3+CaSO_4$

According to stoichiometry :

1 mole of $Ca(NO_3)_2$ require = 1 mole of $Li_2SO_4$

Thus 0.061 moles of $Ca(NO_3)_2$ will require= $\frac{1}{1}\times 0.061=0.061moles$ of $Li_2SO_4$

Thus $Ca(NO_3)_2$ is the limiting reagent as it limits the formation of product and $Li_2SO_4$ is the excess reagent.

As 1 mole of $Ca(NO_3)_2$ give = 1 mole of $CaSO_4$

Thus 0.061 moles of $Ca(NO_3)_2$ give = $\frac{1}{1}\times 0.061=0.061moles$ of $CaSO_4$

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