Question

Determine the mass, in grams, of precipitate that forms when $78.8 \mathrm{~mL}$ of $0.0223 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}$ reacts with excess calcium hydroxide. $\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow$

          Determine the mass, in grams, of precipitate that forms when $78.8 \mathrm{~mL}$ of $0.0223 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}$ reacts
with excess calcium hydroxide.
$\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow$

Added by Ver-Nica G.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Sima Sarker

11/22/2022

Step 1

Given: Volume of Na2CO3 solution = 78.8 mL = 0.0788 L Molar concentration of Na2CO3 = 0.0223 mol/L Moles of Na2CO3 = Volume x Molar concentration Moles of Na2CO3 = 0.0788 L x 0.0223 mol/L Moles of Na2CO3 = 0.001757 mol Show more…

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Determine the mass, in grams, of precipitate that forms when $78.8 \mathrm{~mL}$ of $0.0223 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}$ reacts with excess calcium hydroxide. $\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow$