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irina1246 [14]
3 years ago
14

If you burn 48.6 g of hydrogen and produce 434 g of water, how much oxygen reacted?

Chemistry
1 answer:
GarryVolchara [31]3 years ago
8 0

Answer:

            385.69 g of O₂

Solution:

The Balance Chemical equation for said reaction is as follow;

                                       2 H₂  +  O₂    →   2 H₂O

According to Equation,

        4.032 g ( 2 mol) H₂ reacts to produce  =  36.03 g (2 mol) of H₂O

So,

        48.6 g H₂ on reaction will produce  =  X g of H₂O

Solving for X,

                     X =  (48.6 g × 36.03 g) ÷ 4.032 g

                     X  =  434.29 g of H₂O

It means that the H₂ provided is in Excess. Therefore, the yield of product (H₂O) is being controlled by O₂ (Limiting Reagent).

So, According to Equation,

                    36.03 g (2 mol) H₂O is produced by  =  31.998 g (1 mol) of O₂

So,

                434.29 g of H₂O will be produced by  =  X g of O₂

Solving for X,

                     X =  (434.29 g × 31.998 g) ÷ 36.03 g

                    X  =  385.69 g of O₂

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oksano4ka [1.4K]
Volume = 250 mL in liters : 250 / 1000 => 0.25 L

Molarity = 0.85 M

Number of moles:

n = M x V

n = 0.85 x 0.25

n = 0.21 moles

Choice B
3 0
3 years ago
Consider the equation below. CaCO3(s) <—>CaO(s) + CO2(g) what is the equilibrium constant expression for the given reactio
hichkok12 [17]

Answer:

Answer is: Keq = [CO₂].

Explanation:

Balanced chemical reaction: CaCO3(s) ⇄ CaO(s) + CO₂(g).

The equilibrium constant (Keq) is a ratio of the concentration of the products  to the concentration of the reactants.

Pure liquids (shown in chemical reactions by appending (l) to the chemical formula) and solids (shown in chemical equations by appending (s) to the chemical formula) not go in to he equilibrium constant expression, only gas state (shown in chemical reactions by appending (g) to the chemical formula) reactants and products go in to  the equilibrium constant expression

8 0
3 years ago
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There are some cells that do not have DNA. <br> O True<br> O false
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Answer:

false

Explanation:

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2 years ago
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COCl2(g) decomposes according to the equation above. When pure COCl2(g) is injected into a rigid, previously evacuated flask at
mestny [16]

<u>Answer:</u> The value of K_p for the reaction at 690 K is 0.05

<u>Explanation:</u>

We are given:

Initial pressure of COCl_2 = 1.0 atm

Total pressure at equilibrium = 1.2 atm

The chemical equation for the decomposition of phosgene follows:

                  COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)

Initial:            1                    -         -

At eqllm:       1-x                 x        x

We are given:

Total pressure at equilibrium = [(1 - x) + x+ x]

So, the equation becomes:

[(1 - x) + x+ x]=1.2\\\\x=0.2atm

The expression for K_p for above equation follows:

K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}

p_{CO}=0.2atm\\p_{Cl_2}=0.2atm\\p_{COCl_2}=(1-0.2)=0.8atm

Putting values in above equation, we get:

K_p=\frac{0.2\times 0.2}{0.8}\\\\K_p=0.05

Hence, the value of K_p for the reaction at 690 K is 0.05

3 0
3 years ago
What is the ph of an aqueous solution with the hydronium ion concentration : [h3o+] = 3 x 10-5 m ?
BaLLatris [955]

The concentration of hydrogen can be shown as:

[H+ ] = 3 * 10-5 M

pH can be determined as:

pH = - log [H+ ]

= - log (3 * 10-5)

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Thus the pH of solution is 4.53


8 0
3 years ago
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