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Week 6 - Physical-Science

1) Raziel G. Tiquis is an ABM 11-Compassion student at a weekly physical science class on Fridays and Saturdays from 7:30-9:30 am in March 2022. 2) The written work involves calculating limiting reactants for two chemical reactions. For the first reaction of aluminum and hydrogen bromide, hydrogen bromide is the limiting reactant and the moles of hydrogen formed is 2.98 moles. 3) For the second reaction of silicon and nitrogen, silicon is the limiting reactant and the moles of silicon nitride formed is 7.147 moles. 4) The performance task involves calculating limiting reactants for two food reactions:

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258 views2 pages

Week 6 - Physical-Science

1) Raziel G. Tiquis is an ABM 11-Compassion student at a weekly physical science class on Fridays and Saturdays from 7:30-9:30 am in March 2022. 2) The written work involves calculating limiting reactants for two chemical reactions. For the first reaction of aluminum and hydrogen bromide, hydrogen bromide is the limiting reactant and the moles of hydrogen formed is 2.98 moles. 3) For the second reaction of silicon and nitrogen, silicon is the limiting reactant and the moles of silicon nitride formed is 7.147 moles. 4) The performance task involves calculating limiting reactants for two food reactions:

Uploaded by

Kayla Tiquis
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© © All Rights Reserved
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NAME: Raziel G.

Tiquis DATE: March 25-26, 2022


GRADE & SECTION: ABM 11-Compassion SUBJECT: PHYSICAL SCIENCE
SUBJECT TEACHER: MS. ROSE ANN C. MALUNES SCHEDULE: Fri-Sat 7:30-9:30 am

WEEK 6-QUARTER 3

WRITTEN WORKS:
What’s More:
Activity 1. Limiting Reactants Calculation
Directions: Answer the following questions below. Use three significant figures in your computation
and final answer.
1. Consider the following reaction: 2Al + 6HBr 2 AlBr3 -----------+3H2
a. When 3.22 moles of Al react with 4.96 moles of HBr, how many moles of H2 are formed?
Since 2 mol Al=3 mol𝐻2 , the number of moles of 𝐻2 formed by 3.22 mol Al is
3 mol H
Moles of 𝐻2 = 3.22 mol AlA 2 mol Al2
Moles of 𝐻2 = 4.83 mol
Since 6 mol HBr = 3 mol 𝐻2 , the number of moles of 𝐻2 formed by 4.96 mol HBr is
3 mol H2
Moles of 𝐻2 = 4.96 mol HBrA 6 mol HBr
Moles of 𝐻2 = 2.98 mol
b. What is the limiting reactant?
Since HBr produce less number of moles of 𝐻2
HBr is the limiting reactant
Since HBr is the limiting reactant, the number of moles of 𝐻2 formed is
Moles of 𝐻2 = 2.98 mol
2. Consider the following reaction: 3Si + 2N2 Si3N4
a. When 21.44 moles of Si react with 17.62 moles of N2, how many moles of Si3N4 are formed?
Using Si
Mole ratio = 3 Si: 1 mol 𝑆𝑖3 𝑁4
Using N2
Mole ratio = 2 mol N2 : 1 mol 𝑆𝑖3 𝑁4
Using Si
1 𝑚𝑜𝑙 𝑆𝑖𝑎𝑁𝑎
Moles of SiaNa = 21.44 mol 𝑆𝑖𝐴
3 𝑚𝑜𝑙 𝑆𝑖
Moles if SiaNa = 7.147 mol
Using N2
1 𝑚𝑜𝑙 𝑆𝑖𝑎𝑁𝑎
Moles of SiaNa = 17.62 mol 𝑁𝑎𝐴 2 𝑚𝑜𝑙 𝑁𝑎
Moles if SiaNa = 8.810 mol
b. What is the limiting reactant?
Since Si produce less number of moles of 𝑆𝑖3 𝑁4
Si is the limiting reactant
Since Si limiting reactant ,
Moles if SiaNa = 7.147 mol
PERFORMANCE TASKS:
What I Can Do:
Lab: Limiting Reactants Activity—Datasheet
Question No. 1:
a. How many Guava Jelly (GuJe) can be formed using 5 Guava and 23 Jelly?
As one Guava matches four Jelly
5÷1=5
23 ÷ 4 = 5.75
So, using 5 Guava and 23 Jelly we can make 5 GuJe
b. What is the limiting reactant?
The limiting reactant is Guava
c. What is the excess reactant?
The excess reactant is Jelly
d. How much is left over?
5×4 = 20
23-20 = 3 is the left over
e. Use the balanced equation to answer the following question. One Guava has a mass of 2.0
grams and one Jelly has a mass of 1.5 g. How many Guava Jelly can be made with 12.5 grams
of Guava and 15.0 grams of Jelly?
Guava: 12.5 ÷ 2 = 6.25
Jelly: 15.0 ÷ 1.5 = 10
As one Guava matches four Jelly
Guava: 6.25 ÷ 1 = 6.25
Jelly: 10 ÷ 4 = 2.5
So, using 2 Guava and 8 Jelly we can make 2 Guava Jelly
Question No. 2:
a. How many Hot Pansit can be formed using 10 Pansit and 24 Siling Labuyo?
As two (2) Pancit matches six (6) Siling Labuyo
So, 10 ÷ 2 = 5
And 24 ÷ 6 = 4
So, using 8 Pancit and 24 Siling Labuyo, we can make 4 Hot Pancit
b. What is the limiting reactant?
The limiting reactant is Siling Labuyo
c. What is the excess reactant?
The excess reactant is Pancit
d. How much is left over?
4×2 = 8
8-10 = 2 Pancit
Since there will only 8 Pancit that will be used to make 4 Hot Pancit, there will be 2 left over Pancit
e. Use the balanced equation to answer the following question. One Pansit has a mass of 5.0
grams and one Siling Labuyo has a mass of 1.0 gram. How many Hot Pansit can be made from
40.0 grams of Pansit and 26.0 grams of Siling Labuyo?
40.0 grams ÷ 5.0 grams = 8 Pancit
26.0 grams ÷ 1.0 grams = 26 Siling Labuyo
As two (2) Pancit matches six (6) Siling Labuyo
So, 8 ÷ 2 = 4
And 26 ÷ 6 = 4.33
With this, we can make 4 Hot Pancit

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