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andrey2020 [161]
2 years ago
9

How many grams of CO are needed to react with an excess of Fe 2 O 3 to produce 209.7 g Fe?

Chemistry
1 answer:
gavmur [86]2 years ago
4 0

Answer:

Mass = 157.5 g

Explanation:

Given data:

Mass of CO needed  = ?

Mass of Fe formed = 209.7 g

Solution:

Chemical equation:

3CO + F₂O₃   →       2Fe + 3CO₂

Number of moles of Fe:

Number of moles = mass/ molar mass

Number of moles = 209.7 g/ 55.85 g/mol

Number of moles = 3.75 mol

Now we will compare the moles of iron and carbon monoxide.

                               Fe            :              CO

                                 2            :              3

                                3.75         ;             3/2×3.75 = 5.625 mol

Mass of CO:

Mass = number of moles × molar mass

Mass = 5.625 mol × 28 g/mol

Mass = 157.5 g

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pshichka [43]

Answer:

mole fraction of N_2 O = 0.330

mole of fraction SF_4 = 0.669

PRESSURE OF N_2 O = 39127.053 Pa

pressure of SF_4 = 792126.36

Total pressure   = 118253.413 Pa

Explanation:

Given data:

volume of tank 8 L

Weight of dinitrogen difluoride gas 5.53 g

weight of sulphur hexafluoride gas 17.3 g

Amount of N_2 O = \frac{5.53}{14*2 + 16} = 0.1256 mol

amount of SF_4 = \frac{17.3}{32.1 + 19*4} = 0.254 mol

mole fraction of N_2 O = \frac{0.1256}{0.1256 + 0.254} = 0.330

mole of fractionSF_4 = \frac{0.254}{0.1256 + 0.254} = 0.669

PV = nRT

P of N_2 O = \frac{0.1256 *8.31 (273 + 26.9}{0.008} = 39127.053 Pa

mole of SF_4=\frac{0.254 *8.31*(273+26.9)}{.008} = 79126.36 Pa

Total pressure  = 39127.053 + 79126.36 = 118253.413 Pa

6 0
3 years ago
A sample of hydrogen gas was collected over water. If the total pressure
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mario62 [17]
Scandium has 21 neutrons
6 0
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Nina [5.8K]

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72.0 grams of water how many miles of sodium with react with it?
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Answer:

\large \boxed{\text{8.00 mol}}

Explanation:

We will need a balanced chemical equation with masses, moles, and molar masses.

1. Gather all the information in one place:

Mᵣ:                  18.02

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m/g:                72.0  

2. Moles of H₂O

\text{Moles of H$_{2}$O} = \text{72.0 g H$_{2}$O} \times \dfrac{\text{1 mol H$_{2}$O}}{\text{18.02 g  H$_{2}$O}} = \text{3.996 mol H$_{2}$O}

3. Moles of Na

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7 0
3 years ago
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