Using the balanced reaction below, Given 1.206 g of AICIz,
calculate how many grams of PbCI2 will form
(use periodic table molar masses and choose closest calculated answer below).
3 Pb(NO3)2 (aq) + 2 AIC|3 (ag)
- 3 PbCI2 (s) + 2 Al(NO3)3 (aq)
The molar mass of AICI2 is 118.98 g/mol.
First, we need to convert the mass of AICI2 to moles:
1.206 g / 118.98 g/mol = 0.01014 mol AICI2
From the balanced equation, we know that 3 moles of PbCI2 are produced for every 2 moles of AICI2:
3 mol PbCI2 / 2 mol AICI2
So we can set up a ratio to calculate the moles of PbCI2 produced:
0.01014 mol AICI2 x (3 mol PbCI2 / 2 mol AICI2) = 0.01521 mol PbCI2
Finally, we can convert the moles of PbCI2 to grams using its molar mass:
0.01521 mol PbCI2 x 278.1 g/mol = 4.226 g PbCI2
Rounded to the closest calculated answer below, the answer is 4.23 g PbCI2.