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)?

Feb 14, 2017

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

$\text{A) How many moles of HF are needed to react with 0.260 mol of} N {a}_{2} {O}_{3} S i$

$\text{So for this we need to solve the ratio formula }$

$\text{ The ratio formula is" " :- }$

$1 m o l N {a}_{2} {O}_{3} S i : 8 H F = 1 {H}_{2} S i {F}_{6} : 2 m o l N a F : 3 m o l {H}_{2} O$

So solve for ratio

$\left(0.26 \cdot 1\right) : \left(0.26 \cdot 8\right) = \left(0.26 \cdot 1\right) : \left(0.26 \cdot 2\right) : \left(0.26 \cdot 3\right)$

$\text{0.26mol Na2O3Si : 2.08mol HF = 0.26mol H2SiF6 : 0.52mol NaF : 0.78mol H2O}$

$\text{This means that 2.08 moles of HF are required}$

$\text{B) How many grams of NaF form when 0.600 mol of HF reacts with excess Na_2O_3Si}$

$\text{Since we know already the ratio formula ,lets replace our new }$$\text{variables but this time as we only know the moles of HF we have to use unitary method}$ = $\frac{0.6}{8} \cdot x \text{(variable)}$

$\text{(0.6/8*1molNa2O3Si) :( 0.6molHF) =( 0.6/8*1molH2SiF6): (0.6/8*2molNaF) : (0.6/8*3molH2O)}$

$\text{0.075molNa2O3Si : 0.6molHF = 0.075molH2SiF6 : 0.15molNaF : 0.225molH2O}$

Thus 0.15 moles of NaF are formed.
Mass of 0.15 moles of NaF = $\text{molar mass of NaF * x moles}$
$\text{Molar mass of NaF = 41.98817 g/mol * 0.15 = 6.2982255grams}$

$\text{C)How many grams of Na2O3Si can react with 0.900 g of HF?}$

First we need to find number of moles in 0.900g of HF by using the formula = $\text{Moles" = "grams"/"molar mass}$
Molar mass of HF = 20.01 g per mol
number of moles in 0.900g of HF = ${0.9}_{g} / \text{20.01g per mol} = 0.045 m o l$

the ratio for Na2O3Si to HF = 1 : 8

so we need to use unitary method again

$\text{(0.045/8 * 1mol Na2O3Si) : 0.045mol HF}$

0.00563mol Na2O3Si : 0.045molHF

$\text{Molar mass of Na2O3Si = 122.06 g per mol}$
$\text{Mass of 0.00563mol of NaSiO3 = 0.00563mol * 122.06g per mol = 0.6871978g}$