Question

(Part A) For the reaction, calculate how many grams of the product form when 16.4 g of O2 completely reacts.
Assume that there is more than enough of the other reactant.
4Cr(s)+3O2(g)→2Cr2O3(s)

(Part B) For the reaction, calculate how many grams of the product form when 16.4 g of Sr completely reacts.
Assume that there is more than enough of the other reactant.
2Sr(s)+O2(g)→2SrO(s)

Answer 1

A)

mass of O2 = 16.4 g
molar mass of O2 = 32 g/mol
mol of O2 = (mass)/(molar mass)
= 16.4/32
= 0.5125 mol


According to balanced equation
mol of Cr2O3 formed = (2/3)* moles of O2
= (2/3)*0.5125
= 0.341667 mol



mass of Cr2O3 = number of mol * molar mass
= 0.341667*152
= 51.9 g
Answer: 51.9 g

B)

mass of Sr = 16.4 g
molar mass of Sr = 87.62 g/mol
mol of Sr = (mass)/(molar mass)
= 16.4/87.62
= 0.187172 mol

According to balanced equation
mol of SrO formed = moles of Sr
= 0.187172 mol



mass of SrO = number of mol * molar mass
= 0.187172*103.62
= 19.4 g
Answer: 19.4 g