1) For the following reaction, 26.9 grams of diphosphorus pentoxide are allowed
ID: 1015895 • Letter: 1
Question
1) For the following reaction, 26.9 grams of diphosphorus pentoxide are allowed to react with 9.00 grams of water.
diphosphorus pentoxide (s) + water (l) phosphoric acid(aq)
-What is the maximum amount of phosphoric acid that can be formed? grams
-What is the FORMULA for the limiting reagent?
-What amount of the excess reagent remains after the reaction is complete? grams
2) For the following reaction, 11.6 grams of sulfur are allowed to react with 25.1 grams of carbon monoxide.
sulfur (s) + carbon monoxide (g) sulfur dioxide (g) +carbon (s)
-What is the maximum amount of sulfur dioxide that can be formed? grams
-What is the FORMULA for the limiting reagent?
-What amount of the excess reagent remains after the reaction is complete? grams
3) In the laboratory, a student dilutes 19.2 mL of a 8.02 Mperchloric acid solution to a total volume of 200.0 mL. What is the concentration of the diluted solution?
4) How many milliliters of 9.63 M perchloric acid solution should be used to prepare 1.50 L of 0.100 M HClO4?
5) According to the following reaction:
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (g)
What would you multiply "moles of benzene (C6H6)" by to convert to the units "moles of water" ?
6) According to the following reaction:
SiF4 (s) + 2H2O (l) 4HF (aq) + SiO2 (s)
What would you multiply "grams of silicon tetrafluoride" by to convert to the units "moles of water" ?
7) According to the following reaction, how many moles of waterare necessary to form 0.270 moles magnesium hydroxide?
magnesium nitride (s) + water (l) magnesium hydroxide(aq) + ammonia (aq
8) According to the following reaction, how many moles ofhydrogen gas will be formed upon the complete reaction of 0.433moles water?
water (l) hydrogen (g) + oxygen (g)
9) According to the following reaction, how many moles of waterwill be formed upon the complete reaction of 22.7 grams ofhydrogen peroxide (H2O2)?
hydrogen peroxide (H2O2) (aq) water (l) + oxygen (g)
10) According to the following reaction, how many grams of waterare necessary to form 0.433 moles oxygen gas?
water (l) hydrogen (g) + oxygen (g)
11) In the laboratory a student combines 29.7 mL of a 0.174 Mpotassium nitrate solution with 29.7 mL of a 0.369 Mmagnesium nitrate solution.
What is the final concentration of nitrate anion ?
12) In the laboratory a student combines 49.3 mL of a 0.371 Mchromium(II) bromide solution with 27.5 mL of a 0.484 Mchromium(II) nitrate solution.
What is the final concentration of chromium(II) cation ?
13) A student is asked to standardize a solution of barium hydroxide. He weighs out 1.01 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid).
It requires 28.2 mL of barium hydroxide to reach the endpoint.
A. What is the molarity of the barium hydroxide solution? M
This barium hydroxide solution is then used to titrate an unknown solution of hydroiodic acid.
B. If 15.3 mL of the barium hydroxide solution is required to neutralize 29.7 mL of hydroiodic acid, what is the molarity of thehydroiodic acid solution? M
14) A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.01 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid).
It requires 18.2 mL of potassium hydroxide to reach the endpoint.
A. What is the molarity of the potassium hydroxide solution? M
This potassium hydroxide solution is then used to titrate an unknown solution of nitric acid.
B. If 11.6 mL of the potassium hydroxide solution is required to neutralize 27.5 mL of nitric acid, what is the molarity of thenitric acid solution? M
15) Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solutions. It has the unwieldy formula of KHC8H4O4. This is often written in shorthand notation as KHP.
How many grams of KHP are needed to exactly neutralize 22.8mL of a 0.364 M potassium hydroxide solution ?
16) Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solutions. It has the unwieldy formula of KHC8H4O4. This is often written in shorthand notation as KHP.
-What volume of a 0.163 M potassium hydroxide solution is needed to exactly neutralize 0.888 grams of KHP ?
17) 37.9 mL of 0.201 M hydrobromic acid is added to 31.2 mL ofbarium hydroxide, and the resulting solution is found to be acidic.
18.3 mL of 0.119 M calcium hydroxide is required to reach neutrality.
-What is the molarity of the original barium hydroxide solution?
18) 32.5 mL of 0.913 M perchloric acid is added to 15.1 mL ofcalcium hydroxide, and the resulting solution is found to be acidic.
21.5 mL of 0.479 M barium hydroxide is required to reach neutrality.
-What is the molarity of the original calcium hydroxide solution?
-What is the FORMULA for the limiting reagent?
Explanation / Answer
Solution:- (1) diphosphorus pentoxide is P2O5 and it's reaction with water to give phosphoric acid, H3PO4 is...
P2O5 + 3H2O ----------> 2H3PO4
From this equation, P2O5 and H2O react in 1:3 mol ratio. and the ratio between P2O5 and product is 1:2 where as the ratio between H2O and product is 3:2.
we will calculate the grams of product for the given grams of each reactant and the one that gives less amount of product will be the limiting reactant and the other will be the excess reactant.
26.9 g P2O5 x (1mol/142g) x (2mol H3PO4/1mol P2O5) x (98g/1mol) = 37.1 g H3PO4
9.00 g H2O x (1mol/18g) x (2 mol H3PO4/ 3 mol H2O) x (98g/1mol) = 32.7 g H3PO4
H2O gives maximum (limited amoount) amount of product so it is the limiting reactant and P2O5 is the excess reactant.
So, 32.7 g of H3PO4 can be formed.
Formula of limiting reagent is H2O.
To calculate the excess amount of P2O5 let's calculate the amount of it used to react with given amount of H2O.
9.00 g H2O x (1mol/18g) x (1 mol P2O5 / 3 mol H2O) x (142g/1mol) = 23.7 g P2O5
so, excess amount of P2O5 = 26.9 g - 23.7 g = 3.2 g
hence, the amount of excess reagent aftre the reaction is 3.2 g.
(2) the reaction will be...
S(s) + 2CO(g) -----------> SO2(g) + 2C(s)
We will solve this same as we solved the previous one.
11.6g S x (1mol/32g) x (1 mol SO2 / 1 mol S) x (64 g/1mol) = 23.2 g SO2
25.1 g CO x (1mol/ 28 g) x (1 mol SO2/ 2 mol CO) x (64 g/1mol) = 28.7 g SO2
From calculations, the maximum amount of SO2 that can be formed is 23.2 g.
Formula of the limiting reagent is S.
11.6g S x (1mol/32g) x (2 mol CO/ 1mol S) x (28 g/1mol) = 20.3 g CO
So, only 20.3 g of CO are used to react with given amount of S. amount of excess reagent that is CO after the reaction is = 25.1 g - 20.3 g = 4.8 g
(3) For the dilution problems we could use the dilution equation, M1V1 = M2V2
M1 and M2 are initial and final concentrations. Likewisee V1 and V2 are initial and final volumes. Let's plug in the values...
8.02 M x 19.2 ml = M2 x 200.0 ml
M2 = 8.02 M x 19.2 ml / 200.0 ml
M2 = 0.770 M
So, after dilution the concentration will be 0.770 M.
(4) this one could also be solved same as #3.
9.63 M x V1 = 0.100 M x 1.50 L
V1 = 0.100 M x 1.50 L / 9.63 M = 0.0156 L
convert L into ml.
1 L = 1000 ml
0.0156 L x 1000 ml/ 1L = 15.6 ml
So, the answer is 15.6 ml.
(5) The given balanced equation is...
2C6H6(g) + 15O2(g) ----------> 12CO2(g) + 6H2O(g)
from this equation there is 6:2 that is 3:1 mol ratio between H2O and C6H6. This could also be written as 3mol H2O/1mol C6H6 and it is used to convert moles of C6H6 into moles of H2O.
so, to convert the moles of C6H6 into moles of H2O we need to multiply by 3.
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