PROBLEM 34P

For each of the reactions, calculate how many grams of the product form when 17.8 g of the reactant in color completely reacts. Assume there is more than enough of the other reactant.

(a) Ca(s) + Cl2(g) → CaCl2(s)

(b) 2 K(s) + Br2(l) → 2 KBr(s)

(c) 4 Cr(s) + 3 O2(g) → 2 Cr2O3(s)

(d) 2 Sr(s) + O2(g) → 2 SrO(s)

Solution 34P

Here, we are going to calculate the mass of product when 17.8 g of the reactant in each case completely reacts.

Step 1:

Ca(s) + Cl2(g) → CaCl2(s)Here, the coloured reactant is Chlorine.

Given, mass of Cl2 reacted = 17.8 g

Molar mass of Cl2 = 2 x 35.453 = 70.906 g/mol

Therefore, number of moles of Cl2 = given mass / molar mass

=

= 0.25 mol.

From the equation given above, it is seen that 1 mol of Cl2 produces 1 mol of CaCl2.

Therefore, 0.25 mol of Cl2 will produce 0.25 mol of CaCl2.

Now, 1 mol CaCl2 = molar mass of CaCl2

Therefore, 0.25 mol of CaCl2 = 0.25 x molar mass of CaCl2

= 0.25 x 110.98 g/mol

= 27.7 g

Thus 27.7 g CaCl2 will be produced from reaction (a).

Step 2:

b) 2K(s) + Br2(l) → 2KBr(s)

Here, the coloured reactant is Bromine.

Given, mass of Br2 reacted = 17.8 g

Molar mass of Br2 = 2 x 79.904 = 159.8 g/mol

Therefore, number of moles of Br2 = given mass / molar mass

=

= 0.11 mol.

From the equation given above, it is seen that 1 mol of Br2 produces 2 mol of KBr.

Therefore, 0.11 mol of Br2 will produce (0.11 x 2 = 0.22) mol of KBr.

Now, 1 mol KBr = molar mass of KBr

Therefore, 0.22 mol of KBr = 0.22 x molar mass of KBr

= 0.22 x 119.0 g/mol

= 26.18 g

Thus 26.18 g KBr will be produced from reaction (b).