Question

In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of Na2CO3 was quickly spread on the area and CO2 was released by the reaction by the reaction. Was sufficient NA2CO3 used to neutralize all the acid?

Answer 1

First we must write a balanced chemical equation for this reaction

$Na_2CO_3 _(_a_q_)+ 2HNO_3_(_a_q_) \implies 2NaNO_3_(_s_) + CO_2_(_g_) + H_2O_(_l_)$

The mole ratio for the reaction between $HNO_3$ and $Na_2CO_3$ is 1:2. This means 1 moles of $Na_2CO_3$ will neutralize 2 moles $HNO_3$. Now we find the moles of each reactant based on the mass and molar mass.

$2500g HNO_3 \times \frac{mol}{63.01g\ HNO_3} = 39.67 mol\ HNO_3$

$2000g\ Na_2CO_3 \times \frac{mol}{105.99g \ Na_2CO_3} = 18.87 mol\ Na_2CO_3$

$\frac{18.87 mol Na_2CO_3}{39.67\ HNO_3} = \frac{1 molNa_2CO_3}{2 mol HNO_3}$

The $Na_2CO_3$ was enough to neutralize the acid because 18.87:39.67 is the same as 1:2 mol ratio.