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
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
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