STARTING AMOUNT X Determine the mass in grams of CO, that are produced by the complete reaction of 0.05090 moles of C,H, (toluene) according to the following combustion reaction: ADD FACTOR g CO₂ x( ) 18.02 6.022 x 1023 C₂H₂(1)+ 9 O₂(g) → 7 CO₂(g) + 4H₂O(g) 92.14 g C,H, 0.02908 32.00 1 3.920 mol CO₂ 44.01 g 0₂ 4 0.08908 ANSWER g H₂O 15.68 0.3563 9 0.05090 mol O₂ 2.240 RESET 5 308.1 mol H₂O 7 mol C,H,

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STARTING AMOUNT
X
Determine the mass in grams of CO₂ that are produced by the complete reaction of
0.05090 moles of C,H
(toluene) according to the following combustion reaction:
ADD FACTOR
x( )
18.02
g CO₂
6.022 × 10²3
CH()+ 9 0₂(g) →
92.14
g C₂H₂
0.02908
32.00
1
3.920
mol CO₂
7 CO₂(g) + 4H₂O(g)
g 0₂
44.01
4
0.08908
ANSWER
g H₂O
15.68
0.3563
0.05090
mol O₂
2.240
RESET
3
308.1
mol H₂O
7
mol C₂H₂
Transcribed Image Text:STARTING AMOUNT X Determine the mass in grams of CO₂ that are produced by the complete reaction of 0.05090 moles of C,H (toluene) according to the following combustion reaction: ADD FACTOR x( ) 18.02 g CO₂ 6.022 × 10²3 CH()+ 9 0₂(g) → 92.14 g C₂H₂ 0.02908 32.00 1 3.920 mol CO₂ 7 CO₂(g) + 4H₂O(g) g 0₂ 44.01 4 0.08908 ANSWER g H₂O 15.68 0.3563 0.05090 mol O₂ 2.240 RESET 3 308.1 mol H₂O 7 mol C₂H₂
Expert Solution
Step 1 Introduction

This question belongs to Mole concept.

Given Mass of Toluene = 0.05090g

We know that Molar mass of Toluene = 92 g/mole

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