Chemistry exercises

Question

can someone help me ?

A solid consists of soda ash and calcium nitrate. To 10,0 g of this mixture add 250 ml of 0,5 M sulphuric acid. The gas formed takes up a volume of 2,05 l at 25 °C and a pressure of 1030 hPa. Calculate the percentage composition of the mixture.

A solid consists of 92,5 % sodium hydroxide and 7,5 % calcium hydroxide. 0,235 g of this mixture is reacted with a solution of hydrochloric acid; after the addition of 45,6 ml, the reaction is complete. Calculate the molarity of the hydrogen chloride solution.

thanks in advance

Answer 1

1) First of all, soda ash is the "common name" of the compound calcium carbonate. Thus the reaction will be between that and the sulfuric acid, producing carbon dioxide gas, water, and calcium sulfate. The other part of the mixture, calcium nitrate will not react. So, we can use the give info to calculate the number of moles of CO₂ produced using th eideal gas law, then find the mass of CaCO₃ that would produce that amount of CO₂. This would be the part of the 10.0 g of the mixture, the rest being the calcium nitrate. (Note that the info related to the sulfuric acid is irrelevant.)

CaCO₃ + H₂SO₄ → CO₂ + CaSO₄ + H₂O

PV = nRT solve for n. n = PV/RT

= [(1030 hPa) (100 Pa) (1 atm) (2.05 L) (n K)]/[1.013 x 10⁵Pa) ( 1 hPa) (0.0821 a L) ( 298.15 K)] = 0.0852 mol

(100 g CaCO₃/1 mol CaCO₃) (1 mol CaCO₃/1 mol CO₂) o.0852 mol CO₂ = 8.515 g CaCO₃

So, (8.515 g CaCO₃/10.0 5 g mixture) x 100 = 85.15 or 85.2 % CaCO₃ leaving 14.8 % CaNO₃

2) In this question we have to consider both the NaOH & the Ca(OH)₂ reacting with the HCl independently of one another.

First, we will calculate the mass of each from the data. Next, calculate the number of moles of HCl needed to completely react with each, The total moles of HCl divided by the volume of HCl (in liters) will give the molarity.

0.235 g x 0.925 = 0.217 g NaoH in the mixture, leaving 0.018 g of Ca(OH)₂

NaOH + HCl → NaCl + HOH

(1 mol HCl/1 mol NaOH) (1 mol NaOH/40.0 g NaOH) 0.217 g NaOH = 0.005425 mol HCl used

Ca(OH)₂ + 2 HCl → CaCl₂ + 2 HOH

(2 mol HCl/1 mol Ca(OH)₂) (1 mol Ca(OH)₂/74.1 g Ca(OH)₂) 0.018 g Ca(OH)₂ = 0.000486 mol HCl used

The sum of moles of HCl = 0.005911 mol

M = moles/L = 0.005911/0.0456 L = 0.129 Molar