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A is a solution of H2SO4 containing 4.9gdmcube.b is a solution containing xgdmcube of Na2co3.if 25cmcube of b gives 25.5cmcube of a using the question of the reaction calculate: 1 contraction of a in mol/dmcube 2 concentration of b in mol/dmcube 3 the value of x
Accepted Answer
Here's how to solve this chemistry problem:
1. Concentration of A in mol/dm³:
Molar mass of H₂SO₄: (2
1.01 g/mol) + (32.07 g/mol) + (4
16.00 g/mol) = 98.08 g/mol
Concentration (mol/dm³): (4.9 g/dm³) / (98.08 g/mol) = 0.05 mol/dm³
2. Concentration of B in mol/dm³:
Balanced chemical equation: H₂SO₄(aq) + Na₂CO₃(aq) → Na₂SO₄(aq) + H₂O(l) + CO₂(g)
Mole ratio: From the equation, 1 mole of H₂SO₄ reacts with 1 mole of Na₂CO₃.
Moles of H₂SO₄ in 25.5 cm³ of A: (0.05 mol/dm³)
(25.5 cm³ / 1000 cm³/dm³) = 0.001275 mol
Moles of Na₂CO₃ in 25 cm³ of B: 0.001275 mol (same as moles of H₂SO₄ due to the 1:1 ratio)
Concentration of B (mol/dm³): (0.001275 mol) / (25 cm³ / 1000 cm³/dm³) = 0.051 mol/dm³
3. Value of x:
Molar mass of Na₂CO₃: (2
22.99 g/mol) + (12.01 g/mol) + (3
16.00 g/mol) = 105.99 g/mol
Mass of Na₂CO₃ in 25 cm³ of B: (0.051 mol/dm³)
(25 cm³ / 1000 cm³/dm³)
(105.99 g/mol) = 0.135 g
Value of x (g/dm³): (0.135 g / 25 cm³)
(1000 cm³/dm³) = 5.4 g/dm³
Therefore:
Concentration of A: 0.05 mol/dm³
Concentration of B: 0.051 mol/dm³
Value of x: 5.4 g/dm³
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