<span>the balanced equation for the reaction is as follows
Na</span>₂<span>SO</span>₄<span> + BaCl</span>₂<span> ----> 2NaCl + BaSO</span>₄
<span>stoichiometry of Na</span>₂<span>SO</span>₄<span> to BaCl</span>₂<span> is 1:1
first we need to find out which the limiting reactant is
limiting reactant is fully used up in the reaction.
number of Na2So4 moles - 0.5 mol number of BaCl2 moles - 60 g / 208 g/mol = 0.288 mol
since molar ratio is 1:1 equal number of moles of both reactants should react with each other
therefore BaCl2 is the limiting reactant and Na2SO4 is in excess. amount of product formed depends on number of limiting reactant present.
stoichiometry of BaCl</span>₂<span> to BaSO</span>₄<span> is 1:1.
therefore number of BaSO4 moles formed - 0.288 mol</span>
Answer:
731.25 g
Explanation:
The question asks us to calculate the mass of 12.5 moles of NaCl. The individual relative atomic masses of the elements were supplied. We must first obtain the molar mass of sodium chloride as follows;
Molar mass of sodium chloride= 23.0 + 35.5 = 58.5 gmol-1
From the formula;
Number of moles (n) = mass /molar mass
Number of moles of sodium chloride= 12.5 moles
Mass of sodium = The unknown
Molar mass of sodium chloride= 58.5gmol-1
Mass of sodium chloride= number of moles × molar mass
Mass of sodium chloride= 12.5 × 58.5
Mass of sodium chloride= 731.25 g
