Question

(Part A) For the reaction, calculate how many grams of the product form when 16.4 g of Ca completely reacts.
Assume that there is more than enough of the other reactant.
Ca(s)+Cl2(g)→CaCl2(s)

(Part B) For the reaction, calculate how many grams of the product form when 16.4 g of Br2 completely reacts.
Assume that there is more than enough of the other reactant.
2K(s)+Br2(l)→2KBr(s)

Answer 1

A)

mass of Ca = 16.4 g
molar mass of Ca = 40.08 g/mol
mol of Ca = (mass)/(molar mass)
= 16.4/40.08
= 0.409182 mol
  

According to balanced equation
mol of CaCl2 formed = moles of Ca
= 0.409182 mol



mass of CaCl2 = number of mol * molar mass
= 0.409182*110.98
= 45.4 g
Answer: 45.4 g

B)

mass of Br2 = 16.4 g
molar mass of Br2 = 159.8 g/mol
mol of Br2 = (mass)/(molar mass)
= 16.4/159.8
= 0.102628 mol

According to balanced equation
mol of KBr formed = (2/1)* moles of Br2
= (2/1)*0.102628
= 0.205257 mol



mass of KBr = number of mol * molar mass
= 0.205257*119
= 24.4 g
Answer: 24.4 g