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Sholpan [36]
3 years ago
8

2Al + 6HCl --> 2AlCl3 + 3H2 Aluminium reacts with hydrochloric acid. How many grams of aluminum are necessary to produce 11 L

of hydrogen gas at STP? ( 8.8 g )
Chemistry
1 answer:
dem82 [27]3 years ago
7 0

Answer:

8.8g of Al are necessaries

Explanation:

Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.

To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:

<em>Moles H2:</em>

PV = nRT; PV/RT = n

<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>

Replacing:

1atm*11L/0.082atmL/molK*273.15K = n

n = 0.491 moles of H2 must be produced

<em />

<em>Moles Al:</em>

0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required

<em />

<em>Mass Al -Molar mass: 26.98g/mol-:</em>

0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries

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Chromium (Cr) can combine with chlorine (Cl) to form chromium chloride (CrCl3).
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3 0
3 years ago
In the following reaction, how many grams of ammonia (NH3) will react with 27.8 grams of nitric oxide (NO)? 4NH3 + 6NO → 5N2 + 6
Rzqust [24]
<span>4NH</span>₃<span> + 6NO → 5N</span>₂<span> + 6H</span>₂<span>O

mol of NO  =  </span>\frac{mass of NO}{Molar Mass of NO}
 
                  =  \frac{27.8 g}{30.01 g / mol}
                   
                  =  0.93 mol

Based on the balance equation mole ratio of NH₃  :  NO  is   4 : 6
                                                                                            =  2 : 3

If mol  of NO  =  0.93 mol

then mol of NH₃ = \frac{0.93 mol  *  2}{3}
                           
                          =  0.62 mol

Mass of ammonia =  mol  ×  molar mass
          
                             =  0.62 mol   ×  17.03 g/mol
 
                             =  10.54 g

Therefore B is the best answer





3 0
3 years ago
Read 2 more answers
Suppose you have been given the task of distilling a mixture of hexane + toluene. Pure hexane has a refractive index of 1.375 an
miv72 [106K]

Explanation:

The given data is as follows.

         Refractive index of mixture = 1.456

        Refractive index of hexane = 1.375

        Refractive index of toulene = 1.497

Let mole fraction of hexane = x_{1}

and, mole fraction of toulene = x_{2}

Also,        x_{1} + x_{2} = 1

or,            x_{1} = 1 - x_{2}

Hence, calculate the mole fraction of hexane as follows.

        refractive index mixture= mole fraction hexane × ref index hexane + mole fraction toluene × ref index toluene.

              1.456 = (1 - x_{2}) \times 1.375 + x_{2} \times 1.497

               1.456 = 1.375 - 1.375x_{2} + 1.497x_{2}

               0.081 = 0.122x_{2}

                   x_{2} = \frac{0.081}{0.122}

                                = 0.66

Since,     x_{1} = 1 - x_{2}

                          = 1 - 0.66

                          = 0.34

Thus, we can conclude that mole fraction of hexane in your sample is 0.34.        

3 0
3 years ago
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