When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0
1 answer:
Answer:
The mass of PbSO4 formed 15.163 gram
Explanation:
mole of Pb(NO₃)₂ = 1.25 x 0.05 = 0.0625
mole of Na₂SO₄ = 2 x 0.025 = 0.05
Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2 NaNO₃
( Mole/Stoichiometry )
= 0.0625 = 0.05
From (Mole/ Stoichiometry ) we can conclude that Na₂SO₄ is limiting reagent.
Mass of PbSO₄ precipitate = 0.05 x Molecular mass of PbSO₄
= 0.05 x 303.26 g
= 15.163 g
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<h3>Definition of ideal gas</h3>Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
<h3>Ideal gas law</h3>An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of gases:
P×V = n×R×T
<h3>Moles of acetylene</h3>In this case, you know:
- P= 1765 kPa= 17.4192 atm (being 101.325 kPa= 1 atm)
- V= 390 L
- n= ?
- R= 0.082
- T= 23.5 °C= 296.5 K (being 0 °C= 273 K)
Replacing in the ideal gas law:
17.4192 atm× 390 L = n×0.082 × 296.5 K
Solving:
<u><em>n= 279.42 moles</em></u>
Finally, there are 279.42 moles of acetylene in the tank.
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