Solve the problems below. You must show all of your work for credit. Work must b

Question

Solve the problems below. You must show all of your work for credit. Work must be neat and clearly labeled. Significant figures apply. Don't forget your units! Place a box around your final answer. How many moles of KBr will be produced from 7 moles of BaBr_2 BaBr_2 + K_2SO_4 right arrow KBr + BaSO_4 A reaction was predicted to produce 32.4 grams of a compound. When the product was measured, there were only 26.1 grams made. What is the percent yield of this reaction How many moles of carbon dioxide is produced when 10.4 mol of tricarbon octahydride gas is burned in excess oxygen A certain reaction has a 73.6% yield. If 53.8 grams of the product were predicted by stoichiometry to be made, what would the actual yield be According to this chemical reaction, how many grams of iron can be produced from 14 moles of hydrogen (consider the other reactant as excess--don't forget to balance first) Fe_3O_4 + H_2 right arrow Fe + H_2O Gold is reacted with chlorine gas according to the reaction: 2 Au + 3 Cl_2 right arrow 2 AuCl_3 Use the data in the table to determine the percent yield of gold chloride.

Explanation / Answer

11) according the reaction goes for completion

BABr2 + K2SO4 = 2 KBr + BASO4

we know that 1 moles gives 2 moles of KBr and hence 7 moles of BaBr2 will give 14 moles of KBr

12) % yield

= (calculate yield/expected yield ) x 100

= (26.1/32.4) x 100

=80.55%

13)

C3H8 = 3(12.01) + 8(1.01) = 44.11 g/mol

hence number of mole of tricarbon octahydride present in the reaction is

=gm weight of C3H8/ equivalent weight

=10.4/44.11

=0.2358 mole

C3H8 + 5O2 = 3CO2 + 4H2O

from above reaction 1 moles give 3 moles of CO2

hence

number of moles of CO2 = 3 x 0.2358 =0.7074 moles

14)

% yield

= (calculate yield/expected yield ) x 100

73.6 = (actual yield / 53.8) x 100

hence actual yield = 39.5968gm

15) balance reaction is

Fe3O4 + 4 H2 = 3 Fe + 4 H2O

hence from reaction 4 mole of hydrogen will give 3 moles of Fe

it means 1 moles of hydrogen will give 0.75 moles of Fe

hence 14 moles of hydrogen will give 10.5 moles

the molar mass of Fe is 55.845 hence total weight of Fe = 10.5 * 55.845 = 586.3725 gm of Fe

16) from the given reaction 2 moles of gold requires 3 moles of Cl2

first let us calculate the moles

Au = 39.4/196.966569 =0.2

Cl2= 19/70.906 =0.2679603982737709 = 0.268

hence from above we can say 0.2 moles will requires 0.3 moles of Cl2 but the actual Cl2 is 0.268.

hence Cl2 will be limiting reagent for the reaction

assuming reaction goes for completion

3 moles of Cl2 will give 2 moles of AuCl3

hence moles of AuCl3 = (2/3) *0.268 =0.1787 moles

therotical yield = moles x molar mass

= 0.1787 *303.325569 = 54.18616018390545 gm

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