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2 years ago

A. Which reactant is the limiting reagent?

Chemistry

1 answer:

Tasya [4]2 years ago

8 0

Answer:

a. Zinc is the limiting reactant.

b. $m_{ZnBr_2}^{by\ Zn}=162.61gZnBr_2$

c. $m_{Br_2}^{leftover}=6.6g$

Explanation:

Hello there!

a. In this case, when zinc metal reacts with bromine, the following chemical reaction takes place:

$Zn+Br_2\rightarrow ZnBr_2$

Thus, since zinc and bromine react in a 1:1 mole ratio, we can compute their reacting moles to identify the limiting reactant:

$n_{Zn}=47.2g*\frac{1mol}{65.38g} =0.722molZn\\\\n_{Br_2}=122g*\frac{1mol}{159.8g} =0.763molBr_2$

Thus, since zinc has the fewest moles we infer it is the limiting reactant.

b. Here, we compute the grams of zinc bromide via both reactants:

$m_{ZnBr_2}^{by\ Zn}=0.722molZn*\frac{1molZnBr_3}{1molZn} *\frac{225.22gZnBr_2}{1molZnBr_2} =162.61gZnBr_2\\\\m_{ZnBr_2}^{by\ Br_2}=0.763molBr_2*\frac{1molZnBr_3}{1molBr_2} *\frac{225.22gZnBr_2}{1molZnBr_2} =171.95gZnBr_2$

That is why zinc is the limiting reactant, as it yields the fewest moles of zinc bromide product.

c. Here, since just 0.722 mol of bromine would react, we compute the corresponding mass:

$m_{Br_2}^{reacted}=0.722molBr_2*\frac{159.8gBr_2}{1molBr_2} =115.4gBr_2$

Thus, the leftover of bromine is:

$m_{Br_2}^{leftover}=122g-115.4g\\\\m_{Br_2}^{leftover}=6.6g$

Best regards!

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For the following reaction, 22.6 grams of nitrogen monoxide are allowed to react with 4.64 grams of hydrogen gas . nitrogen mono

antiseptic1488 [7]

Answer:

- 10.5 g of N₂

- Limiting reagent: NO

- 3.13 g of H₂ remains

Explanation:

First of all we state the reaction: 2NO(g) + 2H₂(g) → 2H₂O(l) + N₂(g)

We need to find out the limiting reactant and the excess reagent

Ratio in the reactants is 2:2. Let's convert the mass to moles:

22.6 g / 30 g/mol = 0.753 moles of NO

4.64 g / 2 g/mol = 2.32 moles of H₂

Certainly the limiting reagent is the NO and the excess reactant is the hydrogen:

- For 0.753 moles of NO, we need 0.753 moles of H₂ (we have 2.32 moles)

- For 2.32 moles of H₂, we need 2.32 moles of NO (and we don't have enough NO, because we only have 0.753 moles)

As the H₂ is the excess reagent, some moles still remains after the reaction is complete → 2.32 mol - 0.753 mol = 1.567 moles

We convert the moles to mass: 1.567 mol . 2g /1mol = 3.13 g of H₂ remains

As the NO is the limiting reagent, we can work with the equation:

We propose this rule of three: 2 moles of NO can produce 1 mol of N₂

Then, 0.753 moles of NO must produce (0.753 . 1) /2 = 0.376 moles of N₂

We convert the moles to mass 0.376 mol . 28 g / 1 mol = 10.5 g

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What is aluminum’s nuclear charge?

natita [175]

Answer:All six of the ions contain 10 electrons in the 1s, 2s, and 2p orbitals, but the nuclear charge varies from +7 (N) to +13 (Al)

Explanation:

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How many cm 3 are in 0.014 in 3? (1 in = 2.54 cm)

pashok25 [27]

Answer:

0.229 cm³.

Explanation:

The following data were obtained from the question:

Volume (in in³) = 0.014 in³

Volume (in cm³) =?

1 in = 2.54 cm

Next, we shall determine a conversion scale to convert from in³ to cm³. This can be obtained as follow:

1 in = 2.54 cm

Therefore,

1 in³ = 2.54³ cm³

1 in³ = 16.387 cm³

Finally, we shall convert 0.014 in³ to cm³. This can be obtained as follow:

1 in³ = 16.387 cm³

Therefore,

0.014 in³ = 0.014 in³ × 16.387 cm³ / 1 in³

0.014 in³ = 0.229 cm³

Thus, 0.014 in³ is equivalent to 0.229 cm³.

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What identifies a flaw in J. J. Thomson’s model of the atom?

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Answer:

The plum-pudding model did not identify a central nucleus as the source of a positive charge.

Explanation:

Sir Joseph John Thomson, most popularly known as J J Thomson. He was a famous scientist who was awarded the Noble prize for Physics for his discovery of the subatomic particle, electron.

He placed his famous model of an atom which is known as plum pudding model. He proposed that the atoms are described as the negative particles that is floating within a soup of the diffuse positive charge.

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2 years ago

15 grams of HCl should theoretically produce 0.42 grams of H2. The reaction actually produced 0.15 grams of H2. What is the perc

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Answer:

35.7%

Explanation:

The percent yield is calculated by the formula:

<em>[(actual yield) / (theoretical yield)] * 100</em>

In this case, the actual yield is 0.15 grams, and the theoretical yield is 0.42 grams. So, putting these values into the equation, we have:

$\frac{0.15}{0.42} *100=0.357*100=35.7$

Thus, the percent yield of $H_2$ is 35.7%.

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