A)Calculate the mass of NaNO3 that heated strongly and produced 4.5 dm3(cubic decimeter) of O2 gas. The chemical equation for a chemical reaction is given below (2)
2NaNO3 ---> 2NaNO2 +O2
b) Calculate the concentration of HCl solution formed by dissolving 200 g HCl in 300 CM3 of water.(2)

a) 34 g

b) 18.33 moldm^-3

Explanation:

a) The equation of the reaction is;

2NaNO3> 2NaNO2 + O2

Let us recall that 1 mole of a gas occupies 22.4 dm^3. Since 1 mole of O2 is produced in the reaction, we can write;

2 moles of NaNO3 yields 22.4 dm^3 of O2

x moles of O2 yields 4.5 dm^3 of O2

x = 2 * 4.5/22.4

x = 0.4 moles of NaNO3

But number of moles = mass/molar mass

molar mass of NaNO3 = 85 g/mol

Mass of NaNO3 = 0.4 moles * 85 g/mol

Mass of NaNO3 = 34 g

b)

Number of moles in 200g of HCl = mass/molar mass

molar mass of HCl = 36.5 g/mol

Number of moles = 200g/36.5 = 5.5 moles

From;

Number of moles = concentration * volume in dm^3

concentration = Number of moles/ volume in dm^3

But volume = 300 cm^3 or 0.3 dm^3

concentration = 5.5/0.3 = 18.33 moldm^-3

not clear