Respuesta :
Answer : The mass of lead (ii) oxide produced by the decomposition of the reaction will be 2.09 grams.
Explanation : Given,
Mass of [tex]PbCO_3[/tex] = 2.50 g
Molar mass of [tex]PbCO_3[/tex] = 267.21 g/mole
Molar mass of [tex]PbO[/tex] = 223.2 g/mole
The given balanced chemical reaction is:
[tex]PbCO_3(s)\rightarrow PbO(s)+CO_2(g)[/tex]
First we have to calculate the moles of [tex]PbCO_3[/tex].
[tex]\text{ Moles of }PbCO_3=\frac{\text{ Mass of }PbCO_3}{\text{ Molar mass of }PbCO_3}=\frac{2.50g}{267.21g/mole}=0.00936moles[/tex]
Now we have to calculate the moles of [tex]PbO[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]PbCO_3[/tex] react to give 1 mole of [tex]PbO[/tex]
So, 0.0936 moles of [tex]PbCO_3[/tex] react to give 0.00936 moles of [tex]PbO[/tex]
Now we have to calculate the mass of [tex]PbO[/tex]
[tex]\text{ Mass of }PbO=\text{ Moles of }PbO\times \text{ Molar mass of }PbO[/tex]
[tex]\text{ Mass of }PbO=(0.00936moles)\times (223.2g/mole)=2.09g[/tex]
Therefore, the mass of lead (ii) oxide produced by the decomposition of the reaction will be 2.09 grams.
