How many grams of water can be produced when 65.5 grams of sodium hydroxide reacts with excess sulfuric acid? (4 points)

Unbalanced equation: H2SO4 + NaOH → H2O + Na2SO4
@pooja195

Question

How many grams of water can be produced when 65.5 grams of sodium hydroxide reacts with excess sulfuric acid? (4 points)

Unbalanced equation: H2SO4 + NaOH → H2O + Na2SO4 
@pooja195

pooja195 · Answer

H2SO4 + 2 NaOH → 2 H2O + Na2SO4 

M(NaOH) = M(Na) + M(O) + M(H) = 23 + 16 + 1 = 40 g/mol 

M(H2O) = 2M(H) + M(O) = (2 x 1) + 16 = 18 g/mol 


From the equation : 

2 x 40g of NaOH -----> 2 x 18g of H2O 

65,5g of NaOH -----> (36 x 65,5) / 80 = 29,5g of H2O

anonymous · Answer

So how have you come to this conclusion?

anonymous · Answer

@pooja195 Are you there?

danjs · Answer

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