500 ml of 0.25 M any to S o4 is added to an aqueous solution of 15 gram of bacl2 resulting in the formation of white precipitate of insoluble baso4 how many moles and how many grams of BAso4 are formed
Answers
The balanced equation for the reaction is Na2SO4 + BaCl2 → BaSO4 + 2NaCl
Moles of sodium sulphate, n = c X V
Where,
c is concentration of sodium sulphate = 0.25M
V is volume of sodium sulphate in L = 0.5L
Therefore,
n = 0.25 X 0.5 = 0.125mol
Moles of BaCl2 taking part in the reaction can be calculated from the mass, m, 15g and molar mass, M, 208g/mol using formula n = m/M
Therefore,
n = 15/208 = 0.072mol
From the balanced equation, Na2SO4 and BaCl2 react in 1:1 ratio.
As moles of BaCl2 are lesser than that of Na2SO4, BaCl2 is the limiting reagent and hence only 0.072 moles of each reactant will take part in the reaction
As 1 mole of BaSO4 is formed per mole of either of the reactant, moles of BaSO4 formed = 0.072mol
Mass of BaSO4 formed = n X M = 0.072 X 233 = 16.776g or 16.8g
✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️
Answer:
The balanced equation for the reaction is Na2SO4 + BaCl2 → BaSO4 + 2NaCl
Moles of sodium sulphate, n = c X V
Where,
c is concentration of sodium sulphate = 0.25M
V is volume of sodium sulphate in L = 0.5L
Therefore,
n = 0.25 X 0.5 = 0.125mol
Moles of BaCl2 taking part in the reaction can be calculated from the mass, m, 15g and molar mass, M, 208g/mol using formula n = m/M
Therefore,
n = 15/208 = 0.072mol
From the balanced equation, Na2SO4 and BaCl2 react in 1:1 ratio.
As moles of BaCl2 are lesser than that of Na2SO4, BaCl2 is the limiting reagent and hence only 0.072 moles of each reactant will take part in the reaction
As 1 mole of BaSO4 is formed per mole of either of the reactant, moles of BaSO4 formed = 0.072mol
Mass of BaSO4 formed = n X M = 0.072 X 233 = 16.776g or 16.8g
Explanation: