Answer to Question #109062 in General Chemistry for Jamie Strong

Question

How many moles of sodium hydroxide are needed to react with 1.5 moles of aluminum sulfate?

How many moles of aluminum hydroxide are produced from 10.5 moles of sodium hydroxide?

If you start with 225 g of sodium hydroxide how many g of sodium sulfate would be produced?

Convert 16.5 g al2(so3)2 to grams of na2so3

You combine 34.5 g al2(so3)2 with excess NaOH how many grams of al (oh)2 would you expect to obtain? After the reaction is complete you measure your collected product as 14.46 g of al(oh)3 . What is the percent yield?

Answer 1

Answers:

2NaOH+Al₂(SO4)₃=Na₂SO₄+2Al(OH)₃

According to the stoichiometry of the reaction there are 2 moles aluminum sulfate per 2 moles of sodium hydroxide.

So, 1.5 moles of sodium hydroxide are needed to react with 1.5 moles of aluminum sulfate.

Also 10.5 moles of aluminum hydroxide are produced from 10.5 moles of sodium hydroxide.

At the same time, according to the stoichiometry of the reaction from 2 moles of sodium hydroxide only 1mol of sodium sulfate will be produced.

n_Na2SO4=n_NaOH/2

n_NaOH=m_NaOH/M_NaOH=225g / (23+16+1)g*mo^l-1=5.625mol

n_Na2SO4=5.625mol/2=2.8125mol

m_Na2SO4=n_Na2SO4*M_Na2SO4=2.8125mol*(2*23+32+16*4)g*mo^l-1=

399.375g~399g

So, if you start with 225 g of sodium hydroxide 399g of sodium sulfate will be produced.

According to the stoichiometry of the reaction from 1 mol aluminum sulfate 1 mol of sodium sulfate is produced.

m_Na2SO4=n_Na2SO4*M_Na2SO4

n_Na2SO4=n_Al2(SO4)3=m_Al2(SO4)3/M_Al2(SO4)3=16.5g / (2*27+3*(32+4*16))g*mo^l-1=0.048mol

m_Na2SO4=0.048mol*(2*23+32+16*4)g*mo^l-1=6.8g

Thus, 16.5 g Al₂(SO₄)₃ converts to 6.8g grams of Na₂SO₄.

According to the stoichiometry of the reaction from 1 mol aluminum sulfate 2 moles of aluminium hydroxyde are produced.

m_Al(OH)3=n_Al(OH)3*M_Al(OH)3

n_Al(OH)3=2*n_Al2(SO4)3

n_Al2(SO4)3=m_Al2(SO4)3/M_Al2(SO4)3=34.5g /(2*27+3*(32+4*16))g*mo^l-1=0.1mol

n_Al(OH)3=2*0.1mol=0.2mol

m_Al(OH)3=0.2mol*(27+3(16+1))g*mo^l-1=15.6g

Thus, when you combine 34.5 g Al₂(SO₄)₃ with excess NaOH you will expect to obtain 15.6g of Al(OH)_3.

(14.46g/15.6g)*100%=92.7%

If after the reaction is complete you measure your collected product as 14.46 g of Al(OH)₃, then the percent yield is 92.7%