what mass of lead (ii) trioxonitrate (v)will be required to yield, 22g of lead, (ii)chloride on the addition of excess potassium chloride solution
chemical reactions are of several types study the following reactions and than indicate which type each belongs(a)the action of heat on lead (ii) trioxonitrate .(b) the interaction of iron and steam in a close solution . (a) the reaction of chlorine on iron (chlorine solution.
To determine the mass of lead (II) trioxonitrate (V) required to yield 22g of lead (II) chloride, we need to consider the stoichiometry of the reaction between lead (II) trioxonitrate (V) and potassium chloride.
The balanced equation for the reaction is as follows:
Pb(NO3)2 + 2KCl → PbCl2 + 2KNO3
From the equation, we can see that one mole of lead (II) trioxonitrate (V) reacts with two moles of potassium chloride to produce one mole of lead (II) chloride.
The molar mass of lead (II) trioxonitrate (V) (Pb(NO3)2) can be calculated as follows:
Pb = 207.2 g/mol (atomic mass)
N = 14.0 g/mol (atomic mass)
O = 16.0 g/mol (atomic mass)
Molar mass of Pb(NO3)2 = (207.2 g/mol) + 2 * [(14.0 g/mol) + (3 * 16.0 g/mol)]
= 207.2 g/mol + 2 * (14.0 g/mol + 48.0 g/mol)
= 207.2 g/mol + 2 * (62.0 g/mol)
= 207.2 g/mol + 124.0 g/mol
= 331.2 g/mol
Now, we can use the molar mass to calculate the amount of lead (II) trioxonitrate (V) needed using the following equation:
moles = mass / molar mass
moles of Pb(NO3)2 = 22g / 331.2 g/mol
= 0.0664 mol
Since the stoichiometry of the reaction is 1:1 between Pb(NO3)2 and PbCl2, the moles of PbCl2 formed will also be 0.0664 mol.
Finally, we can calculate the mass of PbCl2 formed using the molar mass of PbCl2.
molar mass of PbCl2 = Pb (atomic mass) + 2 * Cl (atomic mass)
= 207.2 g/mol + 2 * 35.5 g/mol
= 207.2 g/mol + 71 g/mol
= 278.2 g/mol
mass of PbCl2 = moles * molar mass
= 0.0664 mol * 278.2 g/mol
= 18.47 g
Therefore, the mass of lead (II) trioxonitrate (V) required to yield 22g of lead (II) chloride is approximately 18.47 grams.