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IgorC [24]
3 years ago
11

How many moles of C6H12O6 will be produced if 22 grams of CO2 are reacted

Chemistry
1 answer:
SSSSS [86.1K]3 years ago
5 0

Answer:

60 grams

Explanation:

We have the balanced equation (without state symbols):

6

H

2

O

+

6

C

O

2

→

C

6

H

12

O

6

+

6

O

2

So, we would need six moles of carbon dioxide to fully produce one mole of glucose.

Here, we got  

88

 

g

of carbon dioxide, and we need to convert it into moles.

Carbon dioxide has a molar mass of  

44

 

g/mol

. So here, there exist

88

g

44

g

/mol

=

2

 

mol

Since there are two moles of  

C

O

2

, we can produce  

2

6

⋅

1

=

1

3

moles of glucose  

(

C

6

H

12

O

6

)

.

We need to find the mass of the glucose produced, so we multiply the number of moles of glucose by its molar mass.

Glucose has a molar mass of  

180.156

 

g/mol

. So here, the mass of glucose produced is

1

3

mol

⋅

180.156

 

g

mol

≈

60

 

g

to the nearest whole number.

So, approximately  

60

grams of glucose will be produced.

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3 years ago
Consider the reaction of CaC2 and water to produce CaCO3 and NH3 according to the reaction CaCN2 + 3H2O → CaCO3 + 2 NH3 . How mu
xz_007 [3.2K]

Answer:

16.27 g  of CaCO3 are produced upon reaction of 45 g CaCN2 and 45 g of H2O.

Explanation:

Ca(CN)2 + 3H2O → CaCO3 + 2 NH3

First of all, let's find out the limiting reactant.

Molar mass Ca(CN)2.

Molar mass H2O: 18 g/m

Moles of Ca(CN)2: mass / molar mass

45 g / 92.08 g/m = 0.488 moles

Moles of H2O: mass / molar mass

45g / 18g/m = 2.50 moles

This is my rule of three

1 mol of Ca(CN)2 needs 3 moles of H2O

2.5 moles of Ca(CN)2 needs (2.5 . 3) / 1 = 7.5 moles

I need 7.5 moles of water, but I only have 0.488. Obviously water is the limiting reactant; now we can work on it.

3 moles of water __ makes __ 1 mol of CaCO3

0.488 moles of water __ makes ___ (0.488 . 1) / 3 = 0.163 moles

Molar mass CaCO3 = 100.08 g/m

Molar mass . moles = mass

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4 0
3 years ago
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