. Calculate the mass of salt that can be obtained from the reaction of 240g of sodium hydroxide with an excess amount of sulphuric acid.
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Answer:
392gm
Explanation:
Given,
compounds are - NaOH and H2SO4
NaOH = Na+O+H = 23+16+1 = 40 amu
H2SO4 = (H)2 + S + (O)4 = 2+32+64= 98 amu
Thus,
the reaction =
NaOH + H2SO4 ---------> Na2SO4 + H2O
the no. of moles of NaOH =
weight / molecular.weight = 240/40 = 4 moles.
Since,
the ratio = 1 : 1 between Na2SO4 and NaOH
therefore,
moles = weight /molecular weight
weight = moles × molecular weight = 4× 98 = 392gm.
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