-Formaldehyde (CH2O) is a colorless gas with a characteristic pungent odor. It is 53.29% oxygen by mass. Find the mass of oxygen present in 0.632 kg of formaldehyde.
-A student carried out the following experiment:
AgNO3 (aq) + NaCl (aq) -----> AgCl (s) + NaNO3 (aq)
To perform the experiment, the student mixed 50.0 mL of 2.50 M silver nitrate with excess sodium chloride. How many grams of silver chloride will form from this reaction? What volume of a 2.30 M NaCl solution is reacting?
-What volume (in L) of 2.55 M potassium chloride is required to react with 750 mL of 4.80 M lead (II) nitrate?
-Iron can be produced by a "thermite" reaction according to the following equation
Fe2O3 (s) + 2 Al (s) ---> 2 Fe (s) + Al2O3 (s)
If 100.0 g of iron (III) oxide is combined with 30.0 g of aluminum, 55.3 g of iron is produced. Identify the limiting reactant and calculate the theoretical yield and percent yield. How much excess reactant will be left over?
PLEASE HELP! 10 POINTS FOR BEST! THANK YOU SO MUCH IN ADVANCED!
-A student carried out the following experiment:
AgNO3 (aq) + NaCl (aq) -----> AgCl (s) + NaNO3 (aq)
To perform the experiment, the student mixed 50.0 mL of 2.50 M silver nitrate with excess sodium chloride. How many grams of silver chloride will form from this reaction? What volume of a 2.30 M NaCl solution is reacting?
-What volume (in L) of 2.55 M potassium chloride is required to react with 750 mL of 4.80 M lead (II) nitrate?
-Iron can be produced by a "thermite" reaction according to the following equation
Fe2O3 (s) + 2 Al (s) ---> 2 Fe (s) + Al2O3 (s)
If 100.0 g of iron (III) oxide is combined with 30.0 g of aluminum, 55.3 g of iron is produced. Identify the limiting reactant and calculate the theoretical yield and percent yield. How much excess reactant will be left over?
PLEASE HELP! 10 POINTS FOR BEST! THANK YOU SO MUCH IN ADVANCED!
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1 .632Kg CH2O(1000g/1kg)x(1mol CH2O/ 30.0258 g CH2O)x(1 mol O/ 1 mol CH2O)x(15.999 g O/1 mol O) that is the whole problem set up to get the mass of oxygen just plug them into the calculator
2 50.0mL AgNO3(1L/1000mL)x(2.50 mol AgNO3/1L)x(1 mol AgCl/ 1mol AgNO3)x(143.23g AgCl/1 mol AgCl) so that is the set up convert mL to L then use the mol/L=M then use mole ratio to convert to grams using the balanced eqn
- i believe for the second answer you use M1V1=M2V2 so it would be V2=M1V1/M2 just plug them into the eqn V2= (50x2.50)/(2.30)
3 so here is the balanced eqn 2KCl+ Pb(NO3)2= 2KNO3+ PbCl2 but in this problem i dont think you need it you use this again M1V1=M2V2 V1=? M1=2.55 V2= 750mL( convert to L first) M2=4.80 isolate V1 plug them in and theres the answer
4. all you have to do is convert each of them using mol to mol ration to find the limiting reactant
100.0 g Fe2O3(1mol Fe2O3/159.691 gFe2O3)x(2 mol Fe/1mol Fe2O3)x(55.847 gFe/1mol Fe)= 69.94 g
30.0g Al(1mol Al/26.982 g Al)(2mol Fe/2mol Al)(55.847 gFe/1mol Fe)= 62.1 g Fe
so the limiting reactant would be al because you can make way more iron with Fe2O3, and the theoretical yield is 62.1 g Fe
the percent yield is the actual/ theorectical so it is 55.3g/62.1gx100= 89%
for excess you just subtract 69.94-62.1 and thats the excess of Fe that will be leftover so it will be 7.84 g leftover of Fe
2 50.0mL AgNO3(1L/1000mL)x(2.50 mol AgNO3/1L)x(1 mol AgCl/ 1mol AgNO3)x(143.23g AgCl/1 mol AgCl) so that is the set up convert mL to L then use the mol/L=M then use mole ratio to convert to grams using the balanced eqn
- i believe for the second answer you use M1V1=M2V2 so it would be V2=M1V1/M2 just plug them into the eqn V2= (50x2.50)/(2.30)
3 so here is the balanced eqn 2KCl+ Pb(NO3)2= 2KNO3+ PbCl2 but in this problem i dont think you need it you use this again M1V1=M2V2 V1=? M1=2.55 V2= 750mL( convert to L first) M2=4.80 isolate V1 plug them in and theres the answer
4. all you have to do is convert each of them using mol to mol ration to find the limiting reactant
100.0 g Fe2O3(1mol Fe2O3/159.691 gFe2O3)x(2 mol Fe/1mol Fe2O3)x(55.847 gFe/1mol Fe)= 69.94 g
30.0g Al(1mol Al/26.982 g Al)(2mol Fe/2mol Al)(55.847 gFe/1mol Fe)= 62.1 g Fe
so the limiting reactant would be al because you can make way more iron with Fe2O3, and the theoretical yield is 62.1 g Fe
the percent yield is the actual/ theorectical so it is 55.3g/62.1gx100= 89%
for excess you just subtract 69.94-62.1 and thats the excess of Fe that will be leftover so it will be 7.84 g leftover of Fe