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:

(a)

Step1: Dissociate soluble ionic compounds

$NaHCO_{3(s)}
ightarrow Na^{+}_{(aq)}+HCO^{-}_{3(aq)}$, $HCl_{(aq)}
ightarrow H^{+}_{(aq)}+Cl^{-}_{(aq)}$, $NaCl_{(aq)}
ightarrow Na^{+}_{(aq)}+Cl^{-}_{(aq)}$. Then the ionic equation is $HCO^{-}_{3(aq)}+H^{+}_{(aq)}
ightarrow CO_{2(g)}+H_{2}O_{(l)}$

(b)(i)

Step1: Use the formula $n = cV$

Given $c = 1.0\ mol/dm^{3}$ and $V=40\ cm^{3}=0.04\ dm^{3}$. Then $n(HCl)=c(HCl)\times V(HCl)=1.0\ mol/dm^{3}\times0.04\ dm^{3}=0.04\ mol$

(b)(ii)

Step1: Determine mole - ratio from balanced equation

From $HCO^{-}_{3(aq)}+H^{+}_{(aq)}
ightarrow CO_{2(g)}+H_{2}O_{(l)}$, the mole - ratio of $HCl$ to $CO_{2}$ is $1:1$. So $n(CO_{2}) = n(HCl)=0.04\ mol$.

Step2: Use the molar volume formula at standard conditions

At standard conditions ($STP$), $V_{m}=22.4\ dm^{3}/mol$. Then $V(CO_{2})=n(CO_{2})\times V_{m}=0.04\ mol\times22.4\ dm^{3}/mol = 0.896\ dm^{3}=896\ cm^{3}$

(c)

Step1: Calculate moles of $NaHCO_{3}$

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

Step2: Compare moles of reactants with mole - ratio

The mole - ratio of $NaHCO_{3}$ to $HCl$ is $1:1$. We have $n(HCl) = 0.04\ mol$ and $n(NaHCO_{3})\approx0.0417\ mol$. Since $n(HCl)

Answer:

(a) $HCO^{-}_{3(aq)}+H^{+}_{(aq)}
ightarrow CO_{2(g)}+H_{2}O_{(l)}$
(b)(i) $0.04\ mol$
(b)(ii) $896\ cm^{3}$
(c) $HCl$ is the limiting reagent because the amount of $HCl$ will be used up first as $n(HCl)