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)
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).
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).