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Sphinxa [80]
1 year ago
14

For the following reaction, 22.0 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas . nitrogen mono

xide ( g ) + hydrogen ( g ) nitrogen ( g ) + water ( l ) What is the maximum amount of nitrogen gas that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams
Chemistry
1 answer:
mash [69]1 year ago
5 0

NO is the limiting reagent and 4.34 g is the amount of the excess reagent that remains after the reaction is complete

<h3>What is a limiting reagent?</h3>

The reactant that is entirely used up in a reaction is called as limiting reagent.

The reaction:

2NO(g) +2H_2(g) → N_2 +2H_2O

Moles of nitrogen monoxide

Molecular weight: M_(_N_O_)=30g/mol

n_(_N_O_) =\frac{mass}{molar \;mass}

n_(_N_O_) =\frac{22.0}{30g/mol}

n_(_N_O_) = 0.73 mol

Moles of hydrogen

Molecular weight: M_(_H_2_)=30g/mol

n_(_H_2_) =\frac{mass}{molar \;mass}

n_(_H_2_) =\frac{5.80g}{2g/mol}

n_(_H_2_) = 2.9 mol

Hydrogen gas is in excess.

NO is the limiting reagent.

The amount of the excess reagent remains after the reaction is complete.

n_(_N_2_) = (2.9 mol- 0.73 mol NO x \frac{1 \;mol \;of \;H_2}{2 \;mole \;of \;NO}) x \frac{2g \;of \;H_2}{mole \;of \;H_2}

n_(_N_2_) =4.34 g

Learn more about limiting reagents here:

brainly.com/question/26905271

#SPJ1

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We can use the ideal gas law equation to find the volume occupied by oxygen gas
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Design a test to determine whether thorium-234 also emits particles. First, explain how Rutherford’s experiment measured positiv
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The characteristics of the α and β particles allow to find  the design of an experiment to measure the ²³⁴Th particles is:

  • On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the beta particle emission from ²³⁴Th.
  • The neutrons cannot be detected in this experiment because they have no electrical charge.

In Rutherford's experiment, the positive particles directed to the gold film were measured on a phosphorescent screen that with each arriving particle a luminous point is seen.

The particles in this experiment are α particles that have two positive charge and two no charged is a helium nucleus.

The test that can be carried out is to place a small ours of Thorium in front of a phosphorescent screen and see if it has flashes, with the amount of them we can determine the amount of particle emitted per unit of time.

Thorium has several isotopes, with different rates and types of emission:

  • ²³²Th emits α particles, it is the most abundant 99.9%
  • ²³⁴Th emits β particles, exists in small traces.

In this case they indicate that the material used is ²³⁴Th, which emits β particles that are electrons, the detection of these particles is more difficult since it has one negative charge, it has much lower mass, but they can travel further than the particles α, therefore, for what type of isotope we have, we can start measuring at a small distance and increase the distance until the reading is constant. At this point all the particles that arrive are β, which correspond to ²³⁴Th.

Neutron detection is much more difficult since these particles have no charge and therefore do not interact with electrons and no flashing on the screen is varied.

In conclusion with the characteristics of the α and β particles we can find the design of an experiment to measure the ²³⁴Th particles is:

  • On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the β particle emission from ²³⁴Th.
  • The neutrons cannot be detected in this experiment because they have no electrical charge.

Learn more about radioactive emission here: brainly.com/question/15176980

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