Answer to Question #104240 in General Chemistry for Lauren Barad

Question

Consider the reaction between gold (III) sulfide and hydrogen gas to produce dihydrogen silfide gas and gold, Au2S3 + 3H2 --> 2Au + 3H2S.

a. How many moles of gold (III) sulfide are present in 100.0g.
b. How many moles of hydrogen are contained in 5.00L at STP?
c. Given these two reactants how many moles of dihydrogen sulfide can be produced?
d. How many moles of the excess reagent are present once the reaction is complete?
e. If a student conducts the about reaction and obtains 29.1 grams of gold, what is their % yield?
f. If a student carries out the reaction at room temperature (21deg C) instead of 0 deg C, would this cause their percent yield to change? Explain.

Answer 1

a) $moles\ of\ gold (III) \ sulfide=\dfrac{given\ mass }{molar \ mass}$$=100/490=0.20$

b) one mole of hydrogen contains 22.4 litres of hydrogen at STP

then 5 litres of hydrogen is equivalent to x moles of hydrogen

$x=5/22.4 =0.22 moles$

c) Here hydrogen gas is the limiting reagent

so complete gold (III) sulfide will not be consumed here

moles of dihydrogen sulfide can be produced is equal to moles of hydrogen consumed as stochiometric coefficient of both the compunds are same

hence H₂S formed will be equal to 0.22 moles

d) as the stochiometric coefficients of gold (III) sulfide and hydrogen are in ratio 1:3 and hydrogen being the limiting then the amount of gold (III) sulfide left will be equal to

$0.20-0.22/3=0.127$

e)The stochiometric coefficient of gold and hydrogen are in ratio 2:3 then the mole of gold formed will be equal to 2/3 moles of hydrogen consumed

as hydrogen is limiting reagent here so it is fully consumed

moles of gold formed = (2/3)*0.22 =0.148 moles

expected amount of gold to be formed = moles * molar mass = 0.148*197 = 29.16 grams

amount formed =29 grams

% yield =$\dfrac{amount \ formed }{expected\ amount }*100=\dfrac{29}{29.16}*100=99.45\%$