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bulgar [2K]
3 years ago
11

What is the maximum mass of glucose (C6H12O6) that can be burned in 10 g of oxygen?

Chemistry
1 answer:
Dahasolnce [82]3 years ago
5 0
C_6H_{12}O_6 + 6 \ O_2 \to 6 \ CO_2 + 6 \ H_2O
1 mole of glucose : 6 moles of oxygen

First calculate the number of moles of oxygen in 10 g:
M=32 \ \frac{g}{mol} \\
m=10 \ g \\
n=\frac{10 \ g}{32 \ \frac{g}{mol}}=0.3125 \ mol

1 mole of glucose reacts with 6 moles of oxygen
x moles of glucose reacts with 0.3125 moles of oxygen
x=\frac{1 \ mol \cdot 0.3125 \ mol}{6 \ mol} = \frac{3125}{60000} \ mol= \frac{5}{96} \ mol

Now calculate the mass of 5/96 moles of glucose.
M=180 \ \frac{g}{mol} \\
n=\frac{5}{96} \ mol \\
m=180 \ \frac{g}{mol} \cdot \frac{5}{96} \ mol=\frac{900}{96} \ g=9.375 \ g

The maximum mass of glucose that can be burned in 10 g of oxygen is 9.375 g.
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Hope this helps!
3 0
3 years ago
1. How many grams of B are present in 3.35 grams of boron tribromide ?
Andre45 [30]

Answer:

The answer to your question is:

Explanation:

1. How many grams of B are present in 3.35 grams of boron tribromide ?

________ grams B.

MW BBr₃ = 251 g

                            251 g of BBr₃ ----------------------  11 g of B

                             3.35 g          -----------------------    x

                           x = (3.35 x 11) / 251 = 0.147 g of B

2. How many grams of boron tribromide contain 4.69 grams of Br ?

________grams boron tribromide.

MW BBr₃ = 251g

                             251g of BBr₃ -----------------   80 g of Br

                                   x               ----------------- 4.69 g

                           x = (4.69 x 251)/ 80 = 14.71 g of BBr₃

3. How many grams of N are present in 4.11 grams of nitrogen trifluoride ?

________grams N.

MW NF₃ = 71 g

                               71 g of NF₃    -----------------   14 g of N

                               4.11 g              ----------------     x

                               x = (4.11 x 14) / 71 = 0.81 g of N

4. How many grams of nitrogen trifluoride contain 3.07 grams of F ?

________grams nitrogen trifluoride.

MW NF₃ = 71

                                71 g of NF₃ ---------------------   19 g of F

                                  x               ---------------------   3.07 g

                                 x = (3.07 x 71) / 19 = 11.5 g of NF₃

5.How many grams of Co3+ are present in 1.16 grams of cobalt(III) iodide?

________grams Co3+.

MW CoI₃ = 440 g

                             440 g of CoI₃ ------------------  59 g of Co

                              1.16 g             ------------------   x

                              x = (1.16 x 59) / 440 = 0.16 g of Co

6. How many grams of cobalt(III) iodide contain 2.28 grams of Co3+?

________grams cobalt(III) iodide.

MW CoI₃ = 440 g

                             440 g of CoI₃ ------------------  59 g of Co

                               x                   -----------------   2.28 g of Co⁺³

                              x = (2.28 x 440) / 59

                              x = 17 g of CoI₃

5 0
3 years ago
A compound is found to contain 64.80% carbon. 13.62% hydrogen. And 21.58% oxygen by weight. What is the empirical formula for th
Trava [24]

Answer:

Look at sheet

Explanation:

Ight so like I'm too lazy to type this out just look at my work. Idk why it works I just know it works

7 0
3 years ago
Read 2 more answers
A substance with a definite volume and a definite shape is classified as a what?
sergeinik [125]

Answer:

Solids

Explanation:

It can be found

8 0
3 years ago
HELPPPPP PLEASE
andreyandreev [35.5K]

Answer:

V=1.18 L

Explanation:

Alternative solution:

Calculate how many moles

2.34

g

C

O

2

is

2.34

g

⋅

1

m

o

l

44

g

C

O

2

=

0.053

m

o

l

Simple mole ratios tell us that

1

mol of any gas at STP is

22.4

L

, but since we only have

0.053

m

o

l

, we times

22.4

L

m

o

l

by

0.053

m

o

l

.

22.4

L

m

o

l

⋅

0.053

m

o

l

=

1.18

L

Answer link

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7 0
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