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pickupchik [31]
3 years ago
13

Gaseous ethane ch3ch3 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o . suppose 21

. g of ethane is mixed with 110. g of oxygen. calculate the maximum mass of water that could be produced by the chemical reaction. round your answer to 2 significant digits.
Chemistry
1 answer:
Anna11 [10]3 years ago
6 0

The chemical reaction for this is:

 

2 C2H6 + 7 O2 => 4 CO2 + 6 H2O

 

Solving for CO2 with each reactant will give:

 

21.0 g C2H6 x (1 mol C2H6/30.08 g C2H6) x (6 mol H2O/2 mol C2H6) x (18 g H2O/1 mol H2O) = 37.70 g H2O

 

110 g O2 x (1 mol O2/32.00 g O2) x (6 mol CO2/7 mol O2) x (18 g H2O/1 mol H2O) = 53.04 g H2O

 

Since the amount of H2O in C2H6 is lower therefore C2H6 is the limiting reactant and the maximum amount of water is only 38 g H2O (2 significant digits)


ANswer:

38 g water

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The following unbalanced equation illustrates the overall reaction by which the body utilizes glucose to produce energy: C6H12O6
s344n2d4d5 [400]

Answer:

the conversion factor is f= 6  mol of glucose/ mol of CO2

Explanation:

First we need to balance the equation:

C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)

the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:

f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction

f = 6 moles of CO2 / 1 mol of glucose = 6  mol of glucose/ mol of CO2

f = 6 mol of CO2/ mol of glucose

for example, for 2 moles of glucose the number of moles of CO2 produced are

n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2

3 0
3 years ago
What volume would 3.01•1023 molecules of oxygen gas occupy at STP?
Sliva [168]
First, find moles of oxygen gas: (3.01 x10^23 molec.)/(6.02 x10^23) =0.5mol O2


Second, multiply moles by the standard molar volume of a gas at STP:(0.5mol)(22.4L) = 11.2L O2
6 0
3 years ago
Base your answers on the graph below, which represents uniform cooling of a sample of a pure substance, starting as a gas. Solid
Karolina [17]

Answer:

D & E

Explanation:

I think this is dealing with latent heat and D & E would be the range where you will find solid and liquid phases in equilibrium, cuz it starts as gas at from A to B, B to C is gas and liquid equilibrium, C to D is liquid, D to E solid and liquid, and then E to F is solid.

7 0
3 years ago
What is the value for ∆Soreaction for the following reaction, given the standard entropy values? 2H2S(g) + SO2(g) 3Srhombic(s) +
Mademuasel [1]

Answer: \Delta S^{0} for the reaction is -186.75 J/K

Explanation:

Change in entropy (\Delta S^{0}) for the given reaction under standard condition is given by-

\Delta S^{0}= [3\times S_{rhombic}^{0}_{(s)}]+[2\times S_{H_{2}O}^{0}_{(g)}]-[2\times S_{H_{2}S}^{0}_{(g)}]-[1\times S_{SO_{2}}^{0}_{(g)}]

So \Delta S^{0} = [3\times 31.8 J/K.mol]+[2\times 188.825 J/K.mol]-[2\times 205.79 J/K.mol]-[1\times 248.22 J/K.mol] = -186.75 J/K

5 0
3 years ago
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Fofino [41]

yrufbd brinckerhoff circ six

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