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balandron [24]
3 years ago
11

Anyone? Please help

Chemistry
1 answer:
snow_tiger [21]3 years ago
7 0

Answer:

The limiting reacting is O2

Explanation:

Step 1: data given

Number of moles O2 = 21 moles

Number of moles C6H6O = 4.0 moles

Step 2: The balanced equation

C6H6O + 7O2 → 6CO2 + 3H2O

Step 3: Calculate the limiting reactant

For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed (21 moles).

C6H6O is in excess.

For 7 moles O2 we need 1 mol C6H6O

For 21 moles O2 we'll need 21/7 = 3 moles C6H6O

There will remain 4.0 - 3.0 = 1 mol C6H6O

Step 4: calculate products

For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O

For 21 moles O2 we'll have 6/7 * 21 = 18 moles CO2

For 21 moles O2 we'll have 3/7 * 21 = 9 moles H2O

The limiting reacting is O2

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Consider the reaction Mg₂Si(s) + 4 H₂O(ℓ) → 2 Mg(OH)₂(aq) + SiH₄(g). How many grams of silane gas (SiH₄) are formed if 25.0 g of
Neko [114]

Answer:

10.60 grams of silane gas are formed.

Explanation:

From the reaction:

Mg₂Si(s) + 4H₂O(l) → 2Mg(OH)₂(aq) + SiH₄(g)          

We know that the limiting reactant is Mg₂Si, so to find the mass of SiH₄ formed we need to calculate the number of moles of Mg₂Si:

\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}}

Where:

m: is the mass of Mg₂Si = 25.0 g

M: is the molar mass of Mg₂Si = 76.69 g/mol

\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}} = \frac{25.0 g}{76.69 g/mol} = 0.33 moles

Now, the stoichiometric relation between Mg₂Si and SiH₄ is 1:1 so:

\eta_{Mg_{2}Si} = \eta_{SiH_{4}} = 0.33 moles

Finally, the mass of SiH₄ is:

m_{SiH_{4}} = \eta_{SiH_{4}}*M_{SiH_{4}} = 0.33 moles*32.12 g/mol = 10.60 g

Therefore, 10.60 grams of silane gas are formed.

I hope it helps you!    

3 0
3 years ago
2.38 to the nearest hundred
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Answer:

heyooooooooooooo!!

2.00

hope this helps >3

Explanation:

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