Question

For each of the reactions, calculate the mass (in grams) of the product formed when 15.86 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.

4Cr(s)+3O2(g)−−−−−→2Cr2O3(s)

M= g

2Sr(s)−−−−+O2(g)→2SrO(s)   

M= g

Answer 1

4Cr(s)+3O2(g)−−−−−→2Cr2O3(s)

There is no underline, hence i am solving the problem assuming that 15.86 g of Chromium is present and oxygen is in excess

Molar mass of Chromium = 52 gm/mol

Number of moles of Chromium = 15.86/52 = 0.305 moles

2 moles of Chromium will give 1 mole of Cr2O3

Hence number of moles of Cr2O3 produced = 0.305/2 - 0.1525 moles

Molar mass of Cr2O3 = 2 * 52 + 3 * 16 = 104 + 48 = 152 gm/mol

Amount of Cr2O3 produced = 152 gm/mol * 0.1525 moles = 23.18 gms

2Sr(s) +O2(g)→2SrO(s)   

There is no underline, hence i am solving the problem assuming that 15.86 g of Strontium is present and oxygen is in excess

Molar mass of Strontium = 87.62 gm/mol

Number of moles of Chromium = 15.86/87.62 = 0.181 moles

2 moles of Strontium will give 2 mole of SrO

Hence number of moles of SrO produced = 0.181 moles

Molar mass of SrO = 87.62 + 16 = 103.62 gm/mol

Amount of SrO produced = 103.62 gm/mol * 0.181 moles = 18.755 gms