Hydrofluoric acid, HF(aq), cannot be stored in glass bottles because compounds called silicates in the glass are attacked by the HF(aq). Sodium silicate (Na2SiO3), for example, reacts as follows: Na2SiO3(s)+8HF(aq)→H2SiF6(aq)+2NaF(aq)+3H2O(l)?

A)How many moles of HF are needed to react with 0.260 mol of Na2SiO3?
B) How many grams of NaF form when 0.600 mol of HF reacts with excess Na2SiO3?
C)How many grams of Na2SiO3 can react with 0.900 g of HF?

1 Answer
Feb 14, 2017

Answer:

A) 2.08 moles of HF
B) 6.2982255 grams
C)0.6871978 grams

Explanation:

#Na_2O_3Si(s)+8HF(aq)→H_2SiF_6(aq)+2NaF(aq)+3H_2O(l)#

For all our questions we need to find the mole ratio formula for this reaction that is

#1 : 8 = 1 : 2 : 3#

Our question is

#"A) How many moles of HF are needed to react with 0.260 mol of" Na_2O_3Si#

#"So for this we need to solve the ratio formula "#

#" The ratio formula is" " :- " #

#1molNa_2O_3Si : 8HF = 1H_2SiF_6 : 2molNaF : 3molH_2O#

So solve for ratio

#(0.26*1) : (0.26*8) = (0.26*1) : (0.26*2) : (0.26*3)#

#"0.26mol Na2O3Si : 2.08mol HF = 0.26mol H2SiF6 : 0.52mol NaF : 0.78mol H2O"#

#"This means that 2.08 moles of HF are required"#

#"B) How many grams of NaF form when 0.600 mol of HF reacts with excess Na_2O_3Si"#

#"Since we know already the ratio formula ,lets replace our new "##"variables but this time as we only know the moles of HF we have to use unitary method"# = #0.6/8 * x"(variable)"#

#"(0.6/8*1molNa2O3Si) :( 0.6molHF) =( 0.6/8*1molH2SiF6): (0.6/8*2molNaF) : (0.6/8*3molH2O)"#

#"0.075molNa2O3Si : 0.6molHF = 0.075molH2SiF6 : 0.15molNaF : 0.225molH2O"#

Thus 0.15 moles of NaF are formed.
Mass of 0.15 moles of NaF = #"molar mass of NaF * x moles"#
#"Molar mass of NaF = 41.98817 g/mol * 0.15 = 6.2982255grams"#

#"C)How many grams of Na2O3Si can react with 0.900 g of HF?"#

First we need to find number of moles in 0.900g of HF by using the formula = #"Moles" = "grams"/"molar mass"#
Molar mass of HF = 20.01 g per mol
number of moles in 0.900g of HF = #0.9_g/"20.01g per mol" = 0.045mol#

the ratio for Na2O3Si to HF = 1 : 8

so we need to use unitary method again

#"(0.045/8 * 1mol Na2O3Si) : 0.045mol HF"#

0.00563mol Na2O3Si : 0.045molHF

#"Molar mass of Na2O3Si = 122.06 g per mol"#
#"Mass of 0.00563mol of NaSiO3 = 0.00563mol * 122.06g per mol = 0.6871978g"#