How can I solve for this ? Calculate the mass of the precipitate formed when 2.27 L of 0.0820 M Ba(OH)2 are mixed with 3.06 L of
0.0664 M Na2SO4.
1 answer:
The balanced chemical reaction is:
Na2SO4 + Ba(OH)2 = 2NaOH +BaSO4
We are given the amount of the reactants to be used. These values will be the starting point of our calculations.
0.0820 mol/L Ba(OH)2 (2.27 L solution) = 0.18614 mol Ba(OH)2
0.0664 mol/L Na2SO4 (3.06 L solution) = 0.20318 mol Na2SO4
The limiting reactant is the Ba(OH)2. The amount for this compound will be used.
0.18614 mol Ba(OH)2 (1 mol BaSO4 / 1 mol Ba(OH)2 ) (233.43 g BaSO4 / 1 mol BaSO4)= 43.4507 g BaSO4
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Molar mass:
O₂ = 32.0 g /mol CH₄ = 16.0 g/mol
2 O₂ + CH₄ = CO₂ + 2 H₂O
2 x 32 g O₂ ---------------> 16 g CH₄
100 g O₂ ------------------> ( mass of CH₄)
mass of CH₄ = 100 x 16 / 2 x 32
mass of CH₄ = 1600 / 64
= 25 g of CH₄
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