First we must write a balanced chemical equation for this reaction
[tex]Na_2CO_3 _(_a_q_)+ 2HNO_3_(_a_q_) \implies 2NaNO_3_(_s_) + CO_2_(_g_) + H_2O_(_l_)[/tex]
The mole ratio for the reaction between [tex]HNO_3[/tex] and [tex]Na_2CO_3[/tex] is 1:2. This means 1 moles of [tex]Na_2CO_3[/tex] will neutralize 2 moles [tex]HNO_3[/tex]. Now we find the moles of each reactant based on the mass and molar mass.
[tex]2500g HNO_3 \times \frac{mol}{63.01g\ HNO_3} = 39.67 mol\ HNO_3[/tex]
[tex]2000g\ Na_2CO_3 \times \frac{mol}{105.99g \ Na_2CO_3} = 18.87 mol\ Na_2CO_3[/tex]
[tex]\frac{18.87 mol Na_2CO_3}{39.67\ HNO_3} = \frac{1 molNa_2CO_3}{2 mol HNO_3}[/tex]
The [tex]Na_2CO_3[/tex] was enough to neutralize the acid because 18.87:39.67 is the same as 1:2 mol ratio.
