# Answer in General Chemistry for Bhawani Shankar #149659

Equation for the reaction:

H2SO4 + 2NaOH “to” Na2SO4 + H2O

Assuming that:

“The units for the values for quantities of the reagents is not given, I want to assume that;

2.31mL of water is produced from 7.85mL of H2SO4 and 11.7g of NaOH”

Determining the limiting reagent:

Moles of NaOH present in 11.7g

Moles of NaOH “=dfrac{Mass}{MM}”

“=dfrac{11.7g}{40g/mol} = 0.2925mol”

Moles of water in 2.31g produced ;

“Moles of water =dfrac{2.31g}{18g/mol} = 0.128mol”

Actual moles of NaOH used to form water;

Mole ratio of water:NaOH = 1:2

Thus moles of NaOH use “=dfrac{2}{1}x0.128 = 0.2567moles”

But there were a total of 0.2925 moles of NaOH present. Therefore, not all the NaOH present reacted, suggesting that H2SO4 could have been the limiting reagent.

Theoretical moles of water that would be produced if all the NaOH was reacted would be;

Theoretical Moles of H2O “=dfrac{1}{2}xmoles of NaOH present”

“=dfrac{1}{2} x 0.2925mol = 0.14625mol”

“Percent yield = dfrac{(actual yield)}{(thoretical)} x 100”

% Yield “=dfrac{0.128}{0.14625} x 100”

“%Yield = dfrac{theoretical yield – actual yield}{theoretical yield}”

“= 87.5%” % yield

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