Question

For each of the reactions, calculate the mass (in grams) of the product formed when 15.46...

For each of the reactions, calculate the mass (in grams) of the product formed when 15.46 g of the BOLD reactant completely reacts. Assume that there is more than enough of the other reactant. Express your answer using four significant figures.

Part A

2K(s)+Cl2(g)––––––→2KCl(s)

Answer: m= ? g

Part B.
2K(s)+Br2(l)––––––→2KBr(s)

Answer: m= ? g

Part C

4Cr(s)+3O2(g)––––––→2Cr2O3(s)

Answer: m= ? g

Part D

2Sr(s)–––––+O2(g)→2SrO(s)

Answer: m= ? g

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Answer #1

. Heil in fant At A. Q k(s) + la cgl – - QKCl (s) Potassium Chlorine Peotassium chronite In all parts other rendere . Given wso weight & KBr produced= No.d moles y Atomic wit d kBr = 0.395 mol x 119.0 glmes = 47.005 gm Round off up to a significant fOh m 11C cl weight of strontium= 15.469m So. no. of Moles of sr= Given will Atomic wt of Sr = | S. 93 9% 87.62 grimas = 0.177

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