1.

How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl ?

Answer»

0.044
0.333
0.011
0.029

Solution :`PBO+2HCl rarr PbCl_(2)+H_(2)O`
Molar MASS of PbO `=207+16=223" g mol"^(-1)`
`therefore"6.5 g PbO"=(6.5)/(223)" mole"=0.029" mole"`
Molar mass of HCl = 36.5 g `mol^(-1)`
`therefore"3.2 g HCl"=(3.2)/(36.5)" mole = 0.0877 mole"`
1 mole of PbO reacts with 2 MOLES of HCl
Thus, PbO is the LIMITING REACTANT.
1 mole of PbO produces 1 mole of `PbCl_(2)`
`therefore 0.029` mole of PbO will produce `PbCl_(2)`
= 0.029 mole.


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