1) Convert the following into a balanced equation: when lead (II) nitrate soluti

Question

1) Convert the following into a balanced equation:

when lead (II) nitrate solution is added to sodium chloride solution, solid lead (II) chloride forms and sodium nitrate solution remains. Be sure to include the state of each reactant and product in the equation.

2) Write a balanced equation for the following by inserting the correct coefficient

KOH(aq)+Cu(NO3)2(aq)--->Cu(OH)2(s)+KNO3(aq)

3) Calculate the mass of each product formed when 44.57g of diborane (B2H6) reacts with excess water

B2H6(g)+H2O(l)--->H3BO3(s)+H2(g) (hint balance equation first to solve problem)

(A) mass of H3BO3

(B) mass of H2

4) A mixture of 0.0600g of hydrogen and 0.0297 mol of oxygen in a closed container is sparked to indicate a reaction.

(A) How many grams of water can form?

(B) which reactant is in excess and how many grams of it remain after the reaction?

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Explanation / Answer

1) Pb(NO3)2 (s)+ 2NaCl(aq ) ---------> PbCl2(s) + 2NaNO3(aq),

2) 2KOH (aq) + Cu(NO3)2 (aq)-----------> Cu(OH)2(s) +2 KNO3 (aq)

3) B2H6 moles = 44.57/27.67 = 1.61 moles,

B2H6 + 3H2O ----> 2H3BO3 + 3H2

A) mass of H3BO3 = ( 1.61 x 2 x 61.83) = 199.19 gm

B) H2 produced =( 1.61x3x2) = 9.66 gm

4) 0.06 gm H2 = 0.03 mol H2 , 0.0297 mol O2 ,

A) H2 + 0.5 O2 ------> H2O ,

water mass formed = ( 0.03x18)= 0.54 gm

B) O2 is excess and

Oxygen left = ( 0.0297x32 )-(0.03x0.5x32) = 0.4704 gm

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