For each of the following unbalanced equations, suppose that exactly 15.0g of each reactant is taken. Determine which reactant is limiting, and calculate what mass of each product is expected. assume the limiting reactant is completely consumed.

Question

a)Al+HCL->AlCl3+H2
b)NaOH+CO2->Na2CO3+H2O
c)Pb(NO3)2+HCL->PbCl2+HNO3
d)K+I2->KI

Answer 1

Write the balanced equation.
2Al + 6HCl ==> 2AlCl3 + 3H2

Convert 15.0 g Al to moles.
Convert 15.0 g HCl to moles.
15.0 g Al x (1 mole Al/26 g) = 0.555
15.0 g HCl x (1 mole HCl/36.5 g) = 0.410

Convert moles of each to mols of either product (but stay consistent). Let's choose AlCl3.
0.555 moles Al x (2 moles AlCl3/2 moles Al) = 0.555 x 1/1 = 0.555 moles AlCl3.

0.410 moles HCl x (2 moles AlCl3/6 moles HCl) = 0.410 x 2/6 = 0.137 moles AlCl3.

The SMALLER number of moles produced by the HCl means two things: (a) HCl is the limiting reagent, and (b)0.137 moles AlCl3 is the amount of product formed using 15.0 g of each reactant initially.

Now convert moles AlCl3 to grams.
0.137 moles AlCl3 x (133.5 g AlCl3/1 mole AlCl3) = 18.3 g.

All of the problems are worked this way. Just follow this template. And you need to redo this one from the beginning for two reasons. First, I estimated the atomic masses and molar masses and you need to use the exact values to obtain a correct answer. Second, you need the practice and this can be a good practice problem because you know the answers for each step.