Suppose we now collect hydrogen gas, H₂(g), over water at 21°C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:

$2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)$

what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express your answer in liters.

Answer : The volume of hydrogen gas that will be collected is 1.85 L

Explanation :

First we have to calculate the number of moles of aluminium.

Given mass of aluminium = 1.35 g

Molar mass of aluminium = 27 g/mol

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

$\text{Moles of aluminium}=\frac{1.35g}{27g/mol}=0.05mol$

The given chemical reaction is:

$2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)$

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

Thus, aluminium is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, 0.005 moles of aluminium will produce = $\frac{3}{2}\times 0.05=0.0750mol$ of hydrogen gas

Now we have to calculate the mass of helium gas by using ideal gas equation.

PV = nRT

where,

P = Pressure of hydrogen gas = 743 Torr

V = Volume of the helium gas = ?

n = number of moles of hydrogen gas = 0.075 mol

R = Gas constant = $62.364\text{ L Torr }mol^{-1}K^{-1}$

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

Now put all the given values in above equation, we get:

$743Torr\times V=0.075mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 294K\\\\V=1.85L$

Hence, the volume of hydrogen gas that will be collected is 1.85 L

8 0

3 years ago

Given the formula representing a compound: what is a chemical name of this compound? 2-pentene 2-pentyne 3-pentene 3-pentyne

Aleks [24]

The question is incomplete. Complete question is attached below.
..........................................................................................................................

Correct Answer: <em>Option 1) 2-pentene</em>

Reason:
Following are the IUPAC rules for naming the compound
1) Select the longest carbon chain. In present case longest carbon chain has 5 carbon atom. Hence, it is a pentane derivative.
2) In case of alkene, replace 'e' of alkane by 'ene'
3) Give lowest number to function group. In present case, it is double bond.

Applying above rules, the IUPAC name of compound is 2-pentene

4 0

2 years ago

Which is the correct statement regarding the relative Rf values of the starting methyl benzoate vs the product, methyl m-nitrobe

Marianna [84]

Answer:

1. The product has a higher Rf value on a silica gel TLC plate because it is more polar than the starting methyl benzoate.

2. False

3. True

Explanation:

In chromatography, there is a stationary phase and a mobile phase. The ratio of the distance moved by a component and the distance moved by the solvent gives the retention factor (Rf).

Since silica gel is a polar solvent, it will retain the more polar product methyl m-nitrobenzoate compared to the methyl benzoate starting material.

In comparing the electrophillic aromatic substitution of m-nitrobenzoate and methyl benzoate, we must remember that the presence of electron withdrawing groups (such as -NO2 and -CHO) on the aromatic compound deactivates the compound towards electrophillic aromatic substitution hence, methyl m-nitrobenzoate is less reactive than methyl benzoate in Electrophilic Aromatic Substition and Methyl benzoate is less reactive than benzene in Electrophilic Aromatic Substition

5 0

3 years ago