Question

The equation for the action of heat on calcium nitrate is
Ca(NO3)2= CaO+4NO2+O2
1.how many moles of Nitrogen dioxide are produced when 1 mole of calcium nitrate decomposes.
2.what volume of oxygen at STP will be produced on heating 65.6 gram of calcium nitrate.
3.find out the mass of calcium nitrate required to produce 5 moles of gaseous products
4.find out the mass of calcium oxide formed when 65.6 gram of calcium nitrate is heated
5.find out the mass of calcium nitrate required to produce 44.8 litre of Nitrogen dioxide at STP

Answer 1

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Answer 2

Explanation:

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

1. According to reaction 2 mole of calcium nitrate gives 4 moles of  nitrogen Dioxide.

Then 1 mole of calcium nitrate will give:$\frac{4]{2}\times 1 = 2 moles$ of nitrogen dioxide

2.Number of moles of calcium nitrate =]$\frac{65.6 g}{164 g/mol}=0.4 moles$

According to reaction 2 mole of calcium nitrate gives 1 moles of oxygen gas.

Then 0.4 moles of calcium nitrate will give =$\frac{1}{2}\times 0.4$ moles of oxygen that is 0.2 moles.

At STP, 1 mol of gas occupies =  22.4 L

So,0.2 mol of oxygen will occupy = $0.2\times 22.4 L=4.48 L$

3. According to reaction 2 moles of calcium nitrate produces 5 moles of gaseous products.

Mass of 2 moles of calcium nitrate=$2 mol\times 164 g/mol=328 g$

4..Number of moles of calcium nitrate =]$\frac{65.6 g}{164 g/mol}=0.4 moles$

According to reaction 2 moles of calcium nitrate produces 2 moles of calcium oxide.

Then 0.4 moles of calcium nitrate will = $\frac{2}{2}\times 0.4$ moles of calcium oxide that is 0.4 moles.

Mass of calcium oxide =$0.4\times 56 g/mol=22.4 g$

5.At STP, 1 mol of gas occupies =  22.4 L

So,44.8 L of nitrogen oxide will be occupied by = $\frac{1}{22.4}\time 44.8=2 moles$

According to reaction 4 moles of nitrogen dioxide is produced from 2 moles of calcium nitrate.

Then 2 moles of nitrogen dioxide will be obtained from = $\frac{2}{4}\times 2$ moles of calcium nitrate that is 1 mole.

Mass of 1 mol of calcium nitrate=$1 mol\times 164 g/mol=164 g$