Question

1. (a) a mass of 3.50 g of sodium hydrogen carbonate reacted completely with 40 cm³ of 1.0 mol/dm³ hydrochloric acid. the equation for the reaction is shown: nahco₃(s)+hcl(aq)→nacl(aq)+co₂(g)+h₂o(l) write the ionic equation of the reaction (b) calculate the (i) the number of moles of hcl that reacted, (ii) the volume of co₂ produced. (c) deduce with a reason the limiting reagent

Explanation:

Step1: Write ionic equation

Split strong - electrolytes into ions. $NaHCO_3(s)$ is solid and not split, $HCl(aq)$ splits into $H^+(aq)$ and $Cl^-(aq)$, $NaCl(aq)$ splits into $Na^+(aq)$ and $Cl^-(aq)$. The ionic equation is $HCO_3^-(s)+H^+(aq)
ightarrow CO_2(g)+H_2O(l)$.

Step2: Calculate moles of HCl

Use the formula $n = cV$. Given $c = 1.0\ mol/dm^3$ and $V=40\ cm^3 = 0.04\ dm^3$. Then $n(HCl)=cV = 1.0\ mol/dm^3\times0.04\ dm^3=0.04\ mol$.

Step3: Calculate moles of $NaHCO_3$

The molar - mass of $NaHCO_3$ is $M(NaHCO_3)=23 + 1+12 + 3\times16=84\ g/mol$. Given $m = 3.50\ g$, then $n(NaHCO_3)=\frac{m}{M}=\frac{3.50\ g}{84\ g/mol}\approx0.0417\ mol$.

Step4: Determine limiting reagent

From the balanced equation $NaHCO_3(s)+HCl(aq)
ightarrow NaCl(aq)+CO_2(g)+H_2O(l)$, the mole ratio of $NaHCO_3$ to $HCl$ is $1:1$. Since $n(HCl) = 0.04\ mol$ and $n(NaHCO_3)\approx0.0417\ mol$, $HCl$ is the limiting reagent.

Step5: Calculate volume of $CO_2$

From the balanced equation, the mole ratio of $HCl$ to $CO_2$ is $1:1$. So $n(CO_2)=n(HCl) = 0.04\ mol$. At standard temperature and pressure (STP), $V = n\times V_m$, where $V_m = 22.4\ dm^3/mol$. Then $V(CO_2)=0.04\ mol\times22.4\ dm^3/mol = 0.896\ dm^3 = 896\ cm^3$.

Answer:

(a) $HCO_3^-(s)+H^+(aq)
ightarrow CO_2(g)+H_2O(l)$
(b)(i) $0.04\ mol$
(b)(ii) $896\ cm^3$
(c) The limiting reagent is $HCl$. Reason: The mole - ratio of $NaHCO_3$ to $HCl$ in the reaction is $1:1$. The number of moles of $HCl$ is $0.04\ mol$ and the number of moles of $NaHCO_3$ is approximately $0.0417\ mol$. Since $n(HCl)