Question

Use the following balanced equation:

 

 Na2SiO3 (s) + 8 HF(aq)         →    H2SiF6 (aq) + 2 NaF(aq) + 3 H2O (l)

How many grams of Na2SiO3 can react with 2.800 g of HF

Answer 1

v=m/M

M(HF)=20g/mol

v(HF)=2.800g/20g/mol=140mol

v(Na2SiO3)=n/8(HF)=140mol/8=17.5mol

m=v*M

M(Na2SiO3)=122g/mol

m(Na2SiO3)=122g/mol*17.5mol=2135g