Question

how much calcium oxide is formed when 82 gm of calcium nitrate is heated also find the volume of nitrogen dioxide formed

Answer 1

2ca(no3)2 --> 2cao + 4no2 +o2

No2=4 moles = 22.4 ×4 = 89.6Lt

328 g of ca(no3)2 releases

89.6 Lt of NO2

82 g of ca(no3)3 releases=89.6/328×82= 22.4 lit

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Answer 2

When 82 gm of calcium nitrate is heated, 28g of calcium oxide and 22.4 l of nitrogen dioxide is formed.

Given:

Weight of calcium nitrate = 82 gm

To find:

Weight of CaO and volume of nitrogen dioxide = ?

Calculation:

• The chemical equation involved in the given reaction is as follows:

$2Ca(NO_3)_2\rightarrow2 CaO + 4 NO_2 + O_2$

• In the above reaction, 2 mole of calcium nitrate gives 2 moles of calcium oxide, 4 moles of nitrogen oxide, and 1 mole of oxygen.

• To find the grams and volume, the molecular weight of reactant as well as product must be calculated.
• Accordingly, the molecular mass is as follows:

$Ca(NO_3)_2$ - 164

CaO - 56

• As 2 moles of $Ca(NO3_)_2$ is involved, the molecular mass is 2 × 164 = 328 g.
• As 2 moles of CaO is produced, the molecular mass is 2 × 56 = 112 g.

Weight of Calcium oxide formed:

328 g of $Ca(NO_3)_2$ gives 112 g of CaO

∴ 82 g of $Ca(NO_3)_2$ gives = $\frac{112}{328}\times 82$ = 0.3414 × 82 = 28

Thus 82 g of $Ca(NO_3)_2$ gives 28 g of calcium oxide.

Volume of nitrogen dioxide formed:

328 g of $Ca(NO_3)_2$ gives 4 volume of $NO_2$

At STP, the volume of gas = 22.4 l

∴ Volume of Nitrogen dioxide formed = $\frac{4\times22.4}{328}\times 82$ = $\frac{89.6}{328}\times 82$ = 02.7× 82 = 22.4 l

Conclusion:

Thus it is inferred that 82g of $Ca(NO_3)_2$ on heating gives, 28 g of Calcium Oxide and 22.4 l of nitrogen dioxide.

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