Answer:
1- 3 Moles of CO2
2- 132 g of CO2
3- 105,6 g of CO2
4- Limiting Reagent O2
<u>Products form based on limiting reagent (384g O2) :</u>
CO2: 316,8 g
H2O: 172,8 g
<u>Products form based on C3H8 (132,33 g):</u>
CO2: 396,99 g
H2O: 216,54 g
Explanation:
<u>Atomic Masses:</u>
C: 12
H: 1
O: 16
<u>Molecular weights:</u>
C3H8: 44 g
O2: 32 g
H2O: 18 g
CO2: 44 g
C3H8 + 5 O2⇒ 3 CO2 + 4 H2O
C3H8 (44g)+ O2 (160 g) ⇒ CO2 (132 g) + H2O (72 g)
In 5 moles of O2 are produced 3 moles of CO2, equivalent to 132 g
For 160g of O2 are produced 132 g of CO2, so 128 g of O2
160 g O2 ⇒ 132 g CO2
128 g O2⇒ × = 105,6 g CO2
(128×132÷160= 105,6)
The limiting reagent is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, because the reaction cannot continue without it.
If I have 132,33 g of C3H8 and 384 g O2 we can calculate:
For 44 g of CH3H8 ⇒160 g of O2
With 132,33 g of CH3H8 ⇒ ×= 481,2 g of O2
(132,33×160÷44=481,2)
As this amount exceeds the quantity of O2 that we have, we can assume that the 384 g O2 will be totally consumed.
<u>Calculations of the products formed in base of quantity of O2 (limiting reagent):</u>
160 g O2 ⇒ 132 g CO2
384 g O2⇒ × = 316,8 g CO2
(384×132÷160= 316,8)
160 g O2 ⇒ 72 g H2O
384 g O2⇒ × = 172,8 g H2O
(384×72÷160= 172,8)
<u>Calculations of the products formed in base of quantity of C3H8 (excess reagent):</u>
44 g C3H8 ⇒ 132 g CO2
132,33 g C3H8 ⇒ × = 396,99 g CO2
(132,33×132÷44=396,99)
44 g C3H8 ⇒ 72 g H2O
132,33 g C3H8⇒ × = 216,54 g H2O
(132,33×72÷44= 216,54)
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