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3 years ago

A 0.0795-g sample of magnesium metal reacts with hydrochloric acid to give 88.5 mL of hydrogen gas at 25 degree C and 766 mm Hg.

Chemistry

1 answer:

Trava [24]3 years ago

3 0

Answer:

(a). 0.0037 mol.

(b). 22.42 L

(c). => When the measuring apparatus are not precise.

=> Error due to parallax.

=> Random errors.

Explanation:

Without mincing words let us dive straight into the solution to the question above. So, we are given the following parameters or information which are going to aid in solving this particular Question;

Mass of magnesium given= 0.0795 g, volume of Hydrogen = 88.5 mL = 0.0885 L, temperature = 25 degree C = 273 + 25 = 298K and pressure = 766 mm Hg = 766/760 = 1.00789 atm.

(a). In order to determine the number of moles of Hydrogen atoms,we will be making use of the formula below;

Number of moles = (pressure × Volume)/ gas constant,R × Temperature.

Thus,the number of moles of Hydrogen = 1.00789 × 0.0885 / [( 0.08206 × 298] = 0.0037 mol.

(b). The molar volume at stp can be calculated by using the same formula that was used in (a) above;

Molar Volume at stp = Gas constant,R × temperature/ Pressure = 0.08206 × 273 / 1 atm = 22.42 L.

(c). The major sources of errors are given below:

=> When the measuring apparatus are not precise.

=> Error due to parallax.

=> Random errors.

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Explanation:

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4 years ago

The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor. The product str

iren2701 [21]

Answer:

Explanation:

The first thing to do is to write out the chemical equation showing the reaction between the two chemical compounds.

C2H4 + HBr ------------------------> C2H5Br.

The chemical species/compounds reacted in the ratio of 1 : 1 : 1. That is one mole of C2H4 reacted with one mole of HBr to give one mole of C2H5Br.

So, we are given; C2H5Br = 58.7 mol%, HBr = 18.3 mol%. Thus, for C2H4, we have [ 100 - (58.7 + 18.3)] = 23 mol%. Therefore, the limiting reagent is HBr.

So, if we have 100 mol/s ;

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Therefore, the total value of the flow rate in the feed = (HBr in feed) mol/s + (C2H4 in feed) mol/s = (58.7 + 18.3)mol/s + (58.7 + 23) mol/s =( 77 + 81.7) mol/s = 158.7 mol/s. .

So, 100/158.7 × 30.5 = 192.2.

Also the feed composition in HBr and C2H4 is;

HBr = 77/158.7 = 0.4852, C2H4 = 81.7/158.7 = 0.515.

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3 years ago

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Now we have to calculate the moles of NaBr.

The balanced chemical reaction is,

$Pb(NO_3)_2(aq)+2NaBr(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)$

From the balanced reaction we conclude that

As, 2 moles of $NaNO_3$ react with 2 moles of $NaBr$

So, 2.87 moles of $NaNO_3$ react with 2.87 moles of $NaBr$

Now we have to calculate the mass of NaBr.

$\text{Mass of }NaBr=\text{Moles of }NaBr\times \text{Molar mass of }NaBr$

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