Answer to Question #318768 in General Chemistry for Henry

Question

25 cm³ of a solution containing T grams of K₂CO₃ in 1000 cm³ of solution were completely neutralized by 20.00 cm³ of a solution containing 0.025 mole of H₂SO₄ in 1 dm³ of solution

Calculate the :

(a) concentration of K₂CO₃ solution in moldm^-3

(b) Value of T

(c) mass of K₂SO₄ formed

(d) volume of CO₂ evolved at STP

(e) number of H₂O molecules

Answer 1

Volume of solution = $25 cm^3$ , T gram of $K_2CO_3$ ,Volume of solution = 1000 cm³

(a

(a) Volume of $H_2SO_4 =$ 20 cm³ , Mole of $H_2SO_4=$ 0.025

Reaction between sulphuric acid and potassium carbonate

$K_2CO_3+H_2SO_4 \longrightarrow K_2SO_4+CO_2+H_2O$

1 mole of potassium carbonate reacts with one mole of sulphuric acid so 0.025 mole of sulphuric react with 0.025 mole of potassium carbonate

Using molarity equation

$n_1M_1V_1= n_2M_2V_2$

$0.025\times M_1\times 25= 0.025\times 0.025 \times 20$

M₁= 0.02 M

(b)mole of potassium carbonate = 0.025

Mass of potassium carbonate = T = $0.025 \times 138=3.45 g$

(c) Mass of K₂SO₄ = $174.26 \times 0.025 = 4.3565 g$

(d) Mole of carbon dioxide produced at STP=0.025

Volume of carbon dioxide = $0.025 \times 22.4 =0.56 L$

(e) mole of water molecules produced = 0.025

number of water molecules = $6.023 \times 10^{23}\times 0.025 = 1.5 \times 10^{22}$