Question

can you please answer these two questions

3 A compound of sodium and sulfur is analyzed to contain following mass of each elements. Calculate is the empirical formula. Show all calculation steps by using conversion factor, give calculator value (c.v.) and Final value (f.v) to get full credit. Show steps (1.2,3) (Na=5.891 g and S = 4.109 g) (6 pt Step 1: Step 2: Step 3: Empirical Formula is: 4 In the following chemical reaction. Balance the molecular equation by adding coefficient. Write the net ionic equation by adding phases. Do not write ionic Eq. (4 pts) Molecular Eq.: HNO, (aq) + Al(OH)2 (aq) + Al(NO3), (aq) + H2O (1) Net ionic Eq.

Answer 1

3.

Step 1: Determining no. of moles of Na in the compound

Molar mass of Na = 23 g.mol^-1 and mass of Na = 5.891 g. So, moles of Na = (5.891 / 23) mol = 0.256130 mol (c.v) = 0.256 mol (f.v).

Step 2: Determining no. of moles of S in the compound

Molar mass of S = 32 g.mol^-1 and mass of S = 4.109 g. So, moles of Na = (4.109 / 32) mol = 0.128406 mol (c.v) = 0.128 mol (f.v).

Step 3: Getiing ratio of moles

Moles of Na : Moles of S = 0.256 : 0.128 = 2 : 1

Empirical formula is = Na₂S.

4.

Since on the right hand side there are three nitrate ions in Al(NO₃)₃. So,in the left hand side there will be three HNO₃ and consequently on the right hand side there will be three H₂O.

So, balanced equation,

$3HNO_3(aq)+Al(OH)_3(aq)\rightarrow Al(NO_3)_3(aq)+3H_2O(l)$

For getting ionic equation we have to dissociate all species to cations and anions:

$3H^+(aq)+3NO_3^-(aq)+Al^{3+}(aq)+3OH^-(aq)\rightarrow Al^{3+}(aq)+3NO_3^-(aq)+3H_2O(l)$

So, Net ionic equation is:

$3H^+(aq)+3OH^-(aq)\rightarrow 3H_2O(l)$

i.e, $\mathbf{H^+(aq)+OH^-(aq)\rightarrow H_2O(l)}$