What volume of Nitrogen measured at s. T.P. Will be produced on heating 33.8g of calcium trioxonitrate

9.2288 L

Explanation:

The equation of the reaction is given as;

2 Ca(NO3)2 → 2 CaO + 4 NO2 + O2

From the reaction;

2 mol of Ca(NO3)2 produces 4 mol of NO2

Converting 33.8g to Number of moles;

Number of moles = Mass / Molar mass

Number of moles = 33.8 g / 164.088 g/mol

Number of moles = 0.206 mol

2 = 4

0.206 = x

Solving for x;

x = 0.206 * 4 / 2

x = 0.412 mol

At STP;

1 mol = 22.4 L

0.412 mol = x

Solving for x;

x = 0.412 * 22.4 = 9.2288 L

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2