Answer:
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
