For each of the reactions, calculate the mass (in grams) of the product formed when 15.93 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant. 1) 2K(s)+Cl2(g)/15.93G→2KCl(s)
2) 2K(s)+Br2(l)/15.93→2KBr(s)
3) 4Cr(s)+3O2(g)/15.93→2Cr2O3(s)
4) 2Sr(s)/15.93+O2(g)→2SrO(s)

1. 33.43 g of KCl

2. 23.70 g of KBr

3. 50.45 g of Cr₂O₃

4. 18.82 g of SrO

Explanation:

Molar mass of  the elements and compounds in each of the reactions:

K = 39.0 g, Cl = 35.5 g, KCl = 74.5 g, Br = 80.0 g, KBr = 119.0 g, Cr = 52.0 g, O = 16.0 g, Cr₂O₃ = 152.0 g, Sr = 88.0 g, SrO = 104.0 g

1) 2K(s)+Cl2(g)/15.93G→2KCl(s)

From the mole ratio of the reaction above, 2 moles of K reacts with 1 mole of Cl₂ to give 2 moles of KCl

78.0 g (2 * 39.0 g) of K reacts with 71.0 g (2*35.5) of Cl₂ to produce 149.0 g(2*74.5) of KCl, therefore, Cl₂ is the limiting reactant.

15.93 g of Cl₂ will react to produce (149/71) * 15.93 of KCl = 33.43 g of KCl

2) 2K(s)+Br2(l)/15.93→2KBr(s)

From the mole ratio of the reaction, 2 moles of K reacts with 1 mole of Br₂ to give 2 moles of KBr

78.0 g (2 * 39.0 g) of K reacts with 160.0 g (2*80) of Br₂ to produce 238.0 g(2*119.0) of KBr, therefore, K is the limiting reactant which though is in excess.

15.93 g of Br₂ will react to produce (238/160) * 15.93 of KBr = 23.70 g of KBr

3) 4Cr(s)+3O2(g)/15.93→2Cr2O3(s)

From the mole ratio of the reaction, 4 moles of Cr reacts with 3 moles of O₂ to give 2 moles of Cr₂O₃

208.0 g (4 * 52.0 g) of Cr reacts with 96.0 g (3*2*16) of O₂ to produce 304.0 g (2*152.0) of Cr₂O₃, therefore, O₂ is the limiting reactant.

15.93 g of O₂ will react to produce (304/96) * 15.93 of Cr₂O₃ = 50.45 g of Cr₂O₃

4) 2Sr(s)/15.93+O2(g)→2SrO(s)

From the mole ratio of the reaction, 2 moles of Sr reacts with 1 mole of O₂ to give 2 moles of SrO

176.0 g (2* 88.0 g) of Sr reacts with 32.0 g (2*16) of O₂ to produce 208.0 g (2*104.0) of SrO, therefore, O₂ is the limiting reactant which though is in excess.

15.93 g of Sr will react to produce (208/176) * 15.93 of SrO = 18.82 g of SrO

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

thats how its supposed to be