#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"#