14. Silver() nitrate reacts with sodium sulfide as shown in the following reacti
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Question
14. Silver() nitrate reacts with sodium sulfide as shown in the following reaction. How many grams of silverl) sulfide are formed when 200 g of silver nitrate reacts with 5.0 g of sodium sulfide? Note: this is a limiting reactant problerm You cannot assume that the reaction is balanced (a) 14.6 g (b) 15.8 g (c) 7.92 g (d) 29.2 g (c) 31.6 g Dinitrogen monoxide gas (also known as decomposes as shown in following e gas (also known as nitrous oxide, or laughing gas) is sometimes used as an anesthetic. It is the tollowing reaction to form nitrogen gas and oxygen gas. In a laboratoey experiment, What was the percent yield of when 10.0 g of dinitrogen monoxide decomposed, 3.16 g of oxygen gas were col oxygen gas? Note: you cannot assume that the reaction is balanced (a) 31,6% (b) 15.8% (c) 93.4% (d) 63.2% (c) 86.9% is the molarity of an FeClblop) solution prepared by dissolving 20.0 g of FeCl, in enough water to make 275 ml (d) 2.23 × 103 M (e) 275 x 0-w of solution? (a) 4.48 × 10-4 M (b) 4.48 x 101 M (e) 2.23 M 17. How many milliliters ofo550 M HIag) are needed to react with 15.0 mL of 0.217 MCa(OH)(ag) in the following reaction? Note: you cannot assume that the reaction is balanced HUaq) + Ca(OH)far) H,0() + Caltag) (d) 11.8 mlL (e) 38.0 ml (a) 0.0263 ml (b) 0.169 mlL (c) 5.92 mL 18. Which of the following is probably a non-electrolyte? (c) HC1O (d) CH,CH CH CH CI (c) MgCO (a) NH CI (b) CH,COH 19. Based on reaction types and reaction driving forces, what should the products of the following reaction be? (Note: you must predict the products of the reaction to answer this, but you don't need to balance the equation). AgNO,(aq) + (NEL),POdaq) ? (a) AgNH4s) + (NO,)hPOag) (c) Ag POds)+NH.NO,(ag) (e) there won't be a reaction because there are no driving forces (b) Ag PO ag) + NHINOMs) (d) Ag,POds) + HOU 20. What is/are the driving force(s) for the following reaction? (Note: you must predict the products of the reaction to answer this. You don't nced to balance the equation, just predict the products) HPOday) + NILOH(aq) ? (c) formation of a gas (a) formation of a precipitate (d) both (a) and (b) (b) formation of water (c) all three of (a), (b), and (c)Explanation / Answer
14.
AgNO3 + Na2S = Ag2S + NaNO3
The balanced equation is
2 AgNO3 + Na2S = Ag2S + 2 NaNO3
In the above reaction equation
2 moles of AgNO3 reacts with 1 mol of Na2S to produce 1 mole of AgS
Now,
20 g of AgNO3
Molar mass of AgNO3 = 170 g/mol
So, 170 g of AgNO3 = 1 mol
1 g of AgNO3 = (1/170) mol
20 g of AgNO3 = (20/170) mol = 0.118 mol (0.118/2 = 0.59 mol)
Molar mass of Na2S = 78 g/mol
So, 78 g of Na2S = 1 mol
1 g of Na2S = (1/78) mol
5 g of Na2S = (5/78) mol = 0.064 mol
So, Na2S is the limiting reagent.
Now,
1 mol of Na2S to produce 1 mole of Ag2S
0.064 mol of Na2S to produce 0.064 mole of Ag2S
Molar mass of Ag2S = 140 g/mol
So, 1 mol of Ag2S = 140 g
0.064 mol of Ag2S = 0.064 x 140 g
= 8.96 g
15.
N2O = N2(g) + O2(g)
The balanced equation is
2N2O = 2N2(g) + O2(g)
In the above reaction equation
2 moles of N2O produces 2 mole of N2 and 1 mole of O2
Now,
10 g of N2O
Molar mass of N2O = 44 g/mol
So, 44 g of N2O = 1 mol
1 g of N2O = (1/44) mol
10 g of N2O = (10/44) mol = 0.23 mol
Now,
2 moles of N2O produces 1 mole of O2
1 moles of N2O produces (1/2) mole of O2
0.23 moles of N2O produces (0.23/2) mole of O2
0.23 moles of N2O produces 0.115 mole of O2
Molar mass of O2 = 32 g/mol
So, 1 mol of O2 = 32 g
0.115 mol of O2 = 0.115 x 32 g = 3.68 g (theoritical yield)
% yield = (practical yield / theoretical yield) x 100
= (3.16 / 3.68) x 100
= 86 %
16.
Molar mass of FeCl3 = 162.2 g/mol
So, 162.2 g of FeCl3 = 1 mol
1 g of FeCl3 = (1/162.2) mol
20 g of FeCl3 = (20/162.2) mol = 0.123 mol
Now, Molarity = Moles / Liter
= (0.123 mol) / (0.275 L)
= 0.45 M
= 4.5 x 10-1 M
17.
HI(aq) + Ca(OH)2(aq) = H2O(l) + CaI2(aq)
The balanced equation is
2HI(aq) + Ca(OH)2(aq) = 2H2O(l) + CaI2(aq)
In the above reaction equation,
2 moles of HI reacts with 1 mole of Ca(OH)2
Now,
15 mL of 0.217 M Ca(OH)2
Moles of Ca(OH)2 = 0.217 M x 0.045 L
= 0.009765 moles
So, moles of HI = 2 x 0.009765 moles = 0.01953 moles
Molarity of HI = 0.550 M
So, Volume of HI = (0.01953 moles) / (0.550 M)
= 0.0355 L
= 35.5 mL
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