e)Although they were formerly called the inert gases, at least the heavier eleme

Question

e)Although they were formerly called the inert gases, at least the heavier elements of Group 8 do form relatively stable compounds. For example, xenon combines directly with elemental fluorine at elevated temperatures in the presence of a nickel catalyst. Use table 1 and table 2. Xe(g) + 2 F2(g) XeF4(s) What is the theoretical mass of xenon tetrafluoride that should form when 163 g of xenon is reacted with 164 g of F2? g= What is the percent yield if only 156 g of XeF4 is actually isolated? f)When elemental copper is strongly heated with sulfur, a mixture of CuS and Cu2S is produced, with CuS predominating. Cu(s) + S(s) CuS(s) 2 Cu(s) + S(s) Cu2S(s) What is the theoretical yield of CuS when 31.3 g of Cu(s) is heated with 52.5 g of S? (Assume only CuS is produced in the reaction.) g= What is the percent yield of CuS if only 6.08 of CuS can be isolated from the mixture? % yield= g)The gaseous hydrocarbon acetylene, C2H2, is used in welders' torches because of the large amount of heat released when acetylene burns with oxygen. 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(g) How many grams of oxygen gas are needed for the complete combustion of 250. g of acetylene? g=

Explanation / Answer

Solution:- (e) These are the stoichiometry problems. Balanced equation is given as...

Xe(g) 2F2(g) ------> XeF4(s)

here masses for both the reactants are given so we would convert these into moles and calculate the mass of product for both of these. The theoretical yield would be the least one that we get from these two reactants.

given mass of Xe = 163 g

given mass of F2 = 164 g

163 g Xe x (1mol/131.293g) x (1mol XeF4/1mol Xe) x (207.2836g/1mol) = 257g XeF4

164 g F2 x (1mol/37.9968g) x (1mol XeF4/2mol F2) x (207.2836g/1mol) = 447g XeF4

we get the least amount from Xe(limiting reactant), so the theoretical yield is 257g.

actual yield is given as 156 g.

percent yield = (actual/theoretical)x 100

percent yield = (156/257) x 100 = 60.7%

(f) Given equations are...

Cu(s) + S(s) ---------> CuS(s)

2Cu(s) + S(s) ---------> Cu2S(s)

We would solve this same as we solved the previous one..

31.3 g Cu x (1mol/63.546g) x (1mol CuS/1mol Cu) x (95.611g/1mol) = 47.1 g CuS

52.5 g S x (1mol/32.06g) x (1mol CuS/1mol S) x (95.611 g/1mol) = 157g CuS

Here Cu is the limiting reactant and so the theoretical yield of CuS is 47.1 g.

percent yield of CuS = (6.08 g/47.1 g) x 100 = 12.9%

(g) Balanced equation is......

2C2H2(g) + 5O2(g) ----------> 4CO2(g) + 2H2O(g)

250. g C2H2 x (1mol/26g) x (5mol O2/2mol C2H2) x (32g/1mol) = 769 g O2

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