What mass of nitrogen v oxide which could be obtained by heating 33.3 g of lead trioxonitrate(v)
Please solve it for me
No answer for this particular question
Lead(II) nitrate is an IUPAC name for Pb(NO3)2.
N2O5 is not a product of heating lead nitrate.
And can you please teach me how to solve it thanks
To find the mass of nitrogen dioxide (NO2) that could be obtained from heating lead(II) nitrate (Pb(NO3)2), we need to determine the stoichiometry of the reaction between the two compounds. The balanced equation for this reaction is:
2Pb(NO3)2 -> 2PbO + 4NO2 + O2
From the equation, we can see that for every 2 moles of lead(II) nitrate, 4 moles of nitrogen dioxide are produced.
1. Calculate the amount of lead(II) nitrate present in 33.3 g using its molar mass.
Molar mass of Pb(NO3)2 = (207.2 g/mol) + [(14.0 g/mol) + (3 * 16.0 g/mol)] * 2 = 331.2 g/mol
Moles of Pb(NO3)2 = mass / molar mass = 33.3 g / 331.2 g/mol
2. Calculate the moles of nitrogen dioxide (NO2) produced using the stoichiometry of the balanced equation.
Moles of NO2 = (moles of Pb(NO3)2) * (4 moles of NO2 / 2 moles of Pb(NO3)2)
3. Convert the moles of NO2 to grams using the molar mass of nitrogen dioxide.
Molar mass of NO2 = (14.0 g/mol) + (2 * 16.0 g/mol) = 46.0 g/mol
Mass of NO2 = moles of NO2 * molar mass of NO2
By following these steps, you can calculate the mass of nitrogen dioxide produced.