Awuah29

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How do you guys calculate these problems? I don't get the answers !
Limit reagents problems

How many moles of UO2(NO3)2 can be produced by reaction of 0.35 moles U3O8, 1.8 moles HNO3, and 0.2 moles O2 according to the following reaction?
2 U3O8 + 12 HNO3 + O2 6 UO2(NO3)2 + 6 H2O

a) 1.40
b) 0.80
c) 1.05
d) 0.90
e) 1.20

and

Mercury and sulfur react to form HgS. A 3.0 g sample of mercury is reacted with 1.0 g of sulfur. What weight of what substance remains unreacted?
(Atomic weights: Hg = 200.59, S = 32.06).

a) 0.48 g S
b) 0.12 g Hg
c) 0.68 g Hg
d) 0.52 g S
e) 0.64 g S

thanks for help
 

g3k

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Awuah29 said:
How do you guys calculate these problems? I don't get the answers !
Limit reagents problems

How many moles of UO2(NO3)2 can be produced by reaction of 0.35 moles U3O8, 1.8 moles HNO3, and 0.2 moles O2 according to the following reaction?
2 U3O8 + 12 HNO3 + O2 6 UO2(NO3)2 + 6 H2O

a) 1.40
b) 0.80
c) 1.05
d) 0.90
e) 1.20

and

Mercury and sulfur react to form HgS. A 3.0 g sample of mercury is reacted with 1.0 g of sulfur. What weight of what substance remains unreacted?
(Atomic weights: Hg = 200.59, S = 32.06).

a) 0.48 g S
b) 0.12 g Hg
c) 0.68 g Hg
d) 0.52 g S
e) 0.64 g S

thanks for help
1) 2 U3O8 + 12 HNO3 + O2 -> 6 UO2(NO3)2 + 6 H2O

The limiting reactant is O2. Hence 1 O2 gives 6 UO2(NO3)2. Hence .2 moles of O2 will give 1.2 grams.

2)Hg2 + S2 -> 2 HgS

Hence 1 mole Hg2 reacts with 1mole S2. Implies
401.18 gms of mercury react with 64.12 grams sulfur
Therefore, 3 gram mercury will react with .48 grams of sulfur
1 gram sulfur reacts with 6 grams of mercury
That implies mercury is the limiting reactant here. Hence the remaining sulfur is .52 grams sulfur

Hope I am rite!!!
 

NoBraces

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1) 2 U3O8 + 12 HNO3 + O2 -> 6 UO2(NO3)2 + 6 H2O

HNO3 is the limiting reagent. 2 mol of HNO3 reacts to form 1 mol UO2(NO3)2. The answer is (d) 0.9 (See work below).

0.35 mol U3O8 x [3 mol UO2(NO3)2 / 1 mol U3O8] = 1.05 mol UO2(NO3)2

1.8 mol HNO3 x [1 mol UO2(NO3)2 / 2 mol HNO3] = 0.9 mol UO2(NO3)2 :thumbup:

0.2 mol O2 x [6 mol UO2(NO3)2 / 1 mol O2] = 1.2 mol UO2(NO3)2

2) Arrived at the same answer as previous post.
 

g3k

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oops!! My mistake!! :oops:
 
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Awuah29

Awuah29

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NoBraces said:
1) 2 U3O8 + 12 HNO3 + O2 -> 6 UO2(NO3)2 + 6 H2O

HNO3 is the limiting reagent. 2 mol of HNO3 reacts to form 1 mol UO2(NO3)2. The answer is (d) 0.9 (See work below).

0.35 mol U3O8 x [3 mol UO2(NO3)2 / 1 mol U3O8] = 1.05 mol UO2(NO3)2

1.8 mol HNO3 x [1 mol UO2(NO3)2 / 2 mol HNO3] = 0.9 mol UO2(NO3)2 :thumbup:

0.2 mol O2 x [6 mol UO2(NO3)2 / 1 mol O2] = 1.2 mol UO2(NO3)2

2) Arrived at the same answer as previous post.
why do you guys take the 0.35 mol U3O8 multiply by [3 mol UO2(NO3)2 / 1 mol U3O8]. I don't get this ! Is there another step before that? Where does the 3mol of U02(N03) 2 come from???
thanks
 

g3k

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Awuah29 said:
why do you guys take the 0.35 mol U3O8 multiply by [3 mol UO2(NO3)2 / 1 mol U3O8]. I don't get this ! Is there another step before that? Where does the 3mol of U02(N03) 2 come from???
thanks
0.35 mol U3O8 x [6mol UO2(NO3)2 / 2 mol U3O8] = 1.05 mol UO2(NO3)2
is the same as:
0.35 mol U3O8 x [3 mol UO2(NO3)2 / 1 mol U3O8] = 1.05 mol UO2(NO3)2