# How many grams of solid sodium bicarbonate will be consumed?

Excess $\ce{NaHCO3 (s)}$ is added to 510 mL of $\ce{Cu(NO3)2 (aq)}$ $\pu{0.240M}$ for the reaction $\ce{Cu(NO3)2 (aq) + 2NaHCO3 (s) -> CuCO3 (s) + 2NaNO3 (aq) + H2O (l) + CO2 (g)}$

$$\ce{Cu(NO3)2_{(aq)} + 2NaHCO3_{(s)} -> CuCO3_{(s)} + 2NaNO3_{(aq)} + H2O_{(l)} + CO2_{(g)}}$$ The equation is telling us that for every $1$ mole of $\ce{Cu(NO3)2}$ consumed, there are $2$ moles of $\ce{NaHCO3}$ correspondingly consumed in the reaction.

So, let us first find how many moles of $\ce{Cu(NO3)2}$ is consumed. Here, $$\text{Molarity} = \frac{\text{#Moles of } \ce{Cu(NO3)2}}{\text{Volume of Solution}}\\ \implies \text{#Moles of } \ce{Cu(NO3)2} = 0.240 \text{ mol.L}^{-1}\times 510 \text{ mL} \times \frac{1\text{ L}}{1000 \text{ mL}} = 0.1224 \text{ mol}$$

Now, think.

$$1 \text{ mole } \ce{Cu(NO3)2} \iff 2 \text{ moles } \ce{NaHCO3}\\ 0.1224 \text{ moles } \ce{Cu(NO3)2} \iff 0.2448 \text{ moles } \ce{NaHCO3}$$

$\text{Weight of } \ce{NaHCO3}= 0.2448 \text{ mol} \times 84.007 \text{ g mol}^{-1} = 20.5649136 \text{ g}$

Since $\ce{NaHCO3}$ is added in excess the limiting reagent for this reaction is $\ce{Cu(NO3)2}$. Now we calculate the weight of $\ce{Cu(NO3)2}$ which will provide grams of $\ce{NaHCO3}$ consumed by it.

$$\text{Weight in g} = \frac{\text{Molarity * Mol. Wt * Volume in ml}}{1000}\\ \ \ = \frac{240 \times 188 \times 510 }{1000}\ = 23.0112 \text{ g}\ce{Cu(NO3)2}$$ The given equation is,

$$\ce{Cu(NO3)2_{(aq)} + 2NaHCO3_{(s)} -> CuCO3_{(s)} + 2NaNO3_{(aq)} + H2O_{(l)} + CO2_{(g)}}$$

From this balanced equation, it is clear that 1 mole of $\ce{Cu(NO3)2_{(aq)}}$ can consume 2 moles of $\ce{NaHCO3}$

So we can write, $$1 \text{ mole } \ce{Cu(NO3)2} \iff 2 \text{ moles } \ce{NaHCO3}\\188 \text{g } \ce{Cu(NO3)2} \iff 2 \times 84 \text{ g } \ce{NaHCO3}\\23.0112 \text { g} \ce{Cu(NO3)2} \iff \frac{2 \times 84 \times 23.0112}{188}\ =20.5632 \text{g} \ce{NaHCO3}$$

Hence, we can conclude that 20.5632 g $\ce{NaHCO3}$ will be consumed.