The given reaction is mentioned below
"HCl + Na_2SO_4 \\longrightarrow NaCl + H_2SO_4"
we will write balanced chemical equation for it
"2HCl + Na_2SO_4 \\longrightarrow 2NaCl + H_2SO_4"
here , one mole of "Na_2SO_4" will give 2 mole of NaCl
gram molecular mass of Na2SO4 is equal to = 142 g/mole
gram molecular mass of NaCl = 58.5 g/mole
A.
here , given mass of Na2SO4 is 300 g
so mole of "Na_2SO_4 = \\frac{given mass}{molar mass} = \\frac {300}{142} = 2.113"
as per above balanced equation , 1 mole of Na2SO4 give 2 mole of HCl
so 2.113 mole of Na2SO4 will give "= 2\\times 2.113" mole of NaCl
so, mass of NaCl produced="2\\times2.113\\times58.5=247.22 g" of NaCl
This is the theoretical yield in above case
B.
As it is given that percent yield was 67 %
so let x be the actual yield
now as per question
"67" % of x = 247.22 g
so , x"=247.22 \\times \\frac{100}{67}" = 368.98 g
so,actual theoretical yield was = 368.98 g of NaCl
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