Question

How many grams of H2S are needed to completely consumer 10.00g of NaOH? Please do it out step by step. I am stuck on the mol ratio of A :B.

In a separate trial, 3.87g of H2S are reacted with excess NaOH. If the % yeild of Na2S is 74.3%, how many grams of Na2S were actually obtained from this reaction?

Answer 1

Stoichiometry of balance reaction indicates the relationship between reactant and products of any chemical reaction.

Balanced chemical equation:

A balanced chemical equation is an equation that contains same number of atoms on each side of reaction.

H2S (g) + 2 NaOH (aq) --à Na2S (aq) + 2 H2O (l)

Given that

10.00g of NaOH

Number of moles = amount in g/ molar mass

= 10.00 g NaOH/ 39.99711 g/mol

= 0.25 moles NaOH

Now calculate the moles of H2S as followS:

0.25 moles NaOH* 1 mole H2S/2 mole NaOH

=0.125 moles H2S

Amount in g = number of mole * molar mass

= 0.125 moles H2S*34.08088 g/mol

= 4.26 g H2S

Problem 2 given that

3.87g. of H2S

excess NaOH.

% yeild of Na2S = 74.3%,

Number of moles = 3.87 g H2S/ 34.08088 g/ mole

= 0.114 moles H2S

Number of mole of Na2S = 0.114 moles H2S *1/1

0.114 moles Na2S

Amount in g = number of moles * molar mass

= 0.114 moles Na2S*78.04454 g/mol

= 8.86 g Na2S theoretical yield

% yeild of Na2S = 74.3%= actual yield / theoretical yield *100

74.3%= actual yield / 8.86 g *100

74.3*8.86/100= actual yield

actual yield =6.58 g Na2S