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melamori03 [73]
4 years ago
8

What volume will water vapor form when 40 liters of hydrogen completely reacts with 25 liters of oxygen? Assume the temperature

and pressure remains constant.
Chemistry
1 answer:
Brut [27]4 years ago
7 0
                                                             2H2           +           O2 ---->2H2O  
number of moles in reaction                 2 mol                  1 mol       2 mol
number of liters in the reaction         2*22.4 L                1*22.4 L    2*22.4L
We can see that volumes of the gases are proportional to coefficients in the reaction ( if gases are under the same conditions), so we can write

                              2H2           +           O2 ---->2H2O  
                               2 L                         1 L          2 L
     given                 40 L                      ( 25 L)     40 L

We can see that we have excess of O2,
because if 2 L H2 are needed 1 L O2, then 40 L of H2 are needed 20 L O2.

So, limiting reactant is H2, and we will need to calculate Volume of H2O using H2.
2L H2 give 2L H2O(gas), so 40 L H2 give 40 L H2O.
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When aqueous solutions of K3PO4 and Ba(NO3)2 are combined, Ba3(PO4)2 precipitates. Calculate the mass, in grams, of the Ba3(PO4)
Firdavs [7]

Answer:

Mass of Ba₃(PO₄)₂ = 0.0361 g

Explanation:

Given data:

Volume of Ba(NO₃)₂ = 1.2 mL (1.2 × 10⁻³ L )

Molarity of Ba(NO₃)₂ = 0.152 M

Volume of K₃PO₄ = 4.2 mL (4.2 × 10⁻³ L)

Molarity of K₃PO₄ =  0.604 M

Mass of Ba₃(PO₄)₂ produced = ?

Solution:

Chemical equation:

3Ba(NO₃)₂  + 2K₃PO₄  → Ba₃(PO₄)₂  + 6KNO₃

Number of moles of Ba(NO₃)₂ = Molarity × Volume in litter

Number of moles of Ba(NO₃)₂ = 0.152 M × 1.2 × 10⁻³ L

Number of moles of Ba(NO₃)₂ = 0.182 × 10⁻³ mol

Number of moles of K₃PO₄ = Molarity × Volume in litter

Number of moles of K₃PO₄ = 0.604 M × 4.2 × 10⁻³ L

Number of moles of K₃PO₄ = 2.537 × 10⁻³ mol

Now we will compare the moles of Ba₃(PO₄)₂ with K₃PO₄ and Ba(NO₃)₂ .

              Ba(NO₃)₂        :         Ba₃(PO₄)₂

                   3                :               1

              0.182 × 10⁻³    :              1/3 ×0.182 × 10⁻³ = 0.060 × 10⁻³ mol

                K₃PO₄           :          Ba₃(PO₄)₂

                   2                 :                1

              2.537 × 10⁻³     :               1/2 ×  2.537 × 10⁻³= 1.269 × 10⁻³ mol

The number of moles of Ba₃(PO₄)₂ produced by  Ba(NO₃)₂  are less it will limiting reactant.

Mass of Ba₃(PO₄)₂ = moles × molar mass

Mass of Ba₃(PO₄)₂ = 0.060 × 10⁻³ mol × 601.93 g/mol

Mass of Ba₃(PO₄)₂ = 36.12 × 10⁻³ g

Mass of Ba₃(PO₄)₂ = 0.0361 g

6 0
4 years ago
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