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Zielflug [23.3K]
3 years ago
15

How many grams of CO₂ can be produced from the combustion of 2.76 moles of butane according to this equation: 2 C₄H₁₀(g) + 13 O₂

(g) → 8 CO₂ (g) + 10 H₂O (g)
Chemistry
1 answer:
Vilka [71]3 years ago
5 0

Answer:

485.76 g of CO₂ can be made by this combustion

Explanation:

Combustion reaction:

2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)

If we only have the amount of butane, we assume the oxygen is the excess reagent.

Ratio is 2:8. Let's make a rule of three:

2 moles of butane can produce 8 moles of dioxide

Therefore, 2.76 moles of butane must produce (2.76 . 8)/ 2 = 11.04 moles of CO₂

We convert the moles to mass → 11.04 mol . 44g / 1 mol = 485.76 g

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100.mL of a .795 M solution of KBr is diluted to 500.mL. what is the new concentration of the solution?
kap26 [50]

Answer:

0.159 M

Explanation:

convert from mL to L then use the equation:

M1V1 = M2V2

rearrange to find M2

\frac{M1V1}{V2} = M2

\frac{(0.795 M)(0.100 L)}{0.500 L} = 0.159 M

4 0
2 years ago
Molybdenum has a molar mass of 95.94g/mol. How many molecules of molybdenum are in 150.0 g of molybdenum
Kitty [74]
We are given the molar mass of Molybdenum as 95.94 g/mol. Also, the chemical symbol for Molybdenum is Mo. This question is asking for the amount of molecules of molybdenum in a 150.0 g sample. However, since molybdenum is a metal and it is in the form of solid molybdenum, Mo (s), it is not actual a molecule. A molecule has one or more atom bonded together. We will instead be finding the amount of atoms of Molybdenum present in the sample. To do this we use Avogadro's number, which is the amount of atoms/molecules of a substance in 1 mole of that substance.

150.0 g Mo/ 95.94 g/mol = 1.563 moles of Mo

1.563 moles Mo x 6.022 x 10²³ atoms/mole = 9.415 x 10²³ atoms Mo

Therefore, there are 9.415 x 10²³ atoms of Molybdenum in 150.0 g.
5 0
3 years ago
Can someone answer an easy chemistry question? A chemistry grad student measures the performance of the new pump in his lab. The
ella [17]
This problem requires our calculation to undergo the dimensional analysis approach. In this approach, you disregard the actual quantity and focus on the units of measurement. This helps us know the units of our final answer. 

First, let's ignore 16. Let's focus on converting the units kPa-mm³/s to mJ/s. The unit kPa stands for kiloPascals which is 1000 times greater than 1 Pa. The unit mJ, on the other hand, stands for millijoules, which is 1000 times lesser than Joules. The relationship between the two is that, Joules = Pa × m³. But since we want our final answer to be mJ, that would be equal to Pa×mm³. Since the original unit already contains mm³, all we have to do is convert kPa to Pa. 

16 kPa-mm³/s * (1000 Pa/1 kPa) = 16,000 Pa-mm³/s

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7 0
2 years ago
Read 2 more answers
if 0.40 mol of h2 and .15 mol of o2 were to reat as completely as possible to produce h20, what mass of the reactant would remai
Sveta_85 [38]

Answer:

0.2g

Explanation:

Given parameters:

Number of moles of H₂  = 0.4mol

Number of moles of O₂  = 0.15mol

Unknown:

Mass of reactant that would remain = ?

Solution:

To solve this problem, we need to know the limiting reactant which is the one in short supply in the given reaction.

  The expression of the reaction is :

                2H₂  + O₂  →   2H₂O

                    2 mole of H₂ will combine with 1 mole of O₂

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The given number of oxygen gas is 0.15mole and it is the limiting reactant.

Hydrogen gas is in excess;

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    0.15 mole of oxygen gas will require 0.15 x 2  = 0.3mole of hydrogen gas

Now, the excess mole of hydrogen gas  = 0.4 mole  - 0.3 mole  = 0.1mole

  Mass of hydrogen gas  = number of mole x molar mass

  Molar mass of hydrogen gas  = 2(1) = 2g/mol

   Mass of hydrogen gas  = 0.1 x 2 = 0.2g

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