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3 years ago

Help :) plz :) ........

Chemistry

2 answers:

kherson [118]3 years ago

8 0

Answer:

D. 4:3

Explanation:

The mole ratio is simply the coefficients in a balanced chemical reaction and their relation to each other. For example, the mole ratio between Al and $Al_{2}O_{3}$ is 2:1, and the ratio between $O_{2}$ and $Al_{2}O_{3}$ is 3:2.

Ilya [14]3 years ago

3 0

Answer:

<h2>Answer </h2>

answer

is B

please mark as brilliant

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In the laboratory, hydrogen gas is usually made by the following reaction: Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq) How many liters

IrinaK [193]

<u>Answer:</u> The volume of hydrogen gas collected over water is 2.13 L

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of zinc = 5.566 g

Molar mass of zinc = 65.4 g/mol

Putting values in above equation, we get:

$\text{Moles of zinc}=\frac{5.566g}{65.4g/mol}=0.0851mol$

For the given chemical reaction:

$Zn(s)+2HCl(aq.)\rightarrow H_2(g)+ZnCl_2(aq.)$

As, HCl is present in excess. So, it is considered as an excess reagent.

Zinc is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of zinc produces 1 mole of hydrogen gas.

So, 0.0851 moles of zinc will produce = $\frac{1}[1}\times 0.0851=0.0851mol$ of hydrogen gas

To calculate the volume of hydrogen gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of hydrogen gas = Total atmospheric pressure - vapor pressure of water = (752 - 18.65) mmHg = 733.35 mmHg

V = Volume of the hydrogen gas

n = number of moles of gas = 0.0851 moles

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of hydrogen gas = $21^oC=[21+273]K=294K$

Putting values in above equation, we get:

$733.35mmHg\times V=0.0851mol\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 294K\\\\V=\frac{0.0851\times 62.3637\times 294}{733.35}=2.13L$

Hence, the volume of hydrogen gas collected over water is 2.13 L

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