Question

Use the following balanced reaction:

Na2​CO3​ + Ca(HC2​H3​O2​)2​ yield 2NaHC2​H3​O2​ + CaCO3​

If you have 5.51 moles of Na2CO3 and 6.01 moles of Ca(HC2H3O2)2, how many moles of NaHC2H3O2 will be produced?

Answer 1

We've got the balanced equation

$$N{a_2}C{O_3} + Ca{(H{C_2}{H_3}{O_2})_2} \to 2NaH{C_2}{H_3}{O_2} + CaC{O_3}$$

and we can see that each 1 mole of $N{a_2}C{O_3}$ reacts with 1 mole of $Ca{(H{C_2}{H_3}{O_2})_2}$ , so the limited reagent in our situation is $N{a_2}C{O_3}$ (because $5.51[{\text{mol]}} < 6.01[{\text{mol]}}$) . Next, we can see that each 1 mole of $N{a_2}C{O_3}$ give us 2 moles of $NaH{C_2}{H_3}{O_2}$ , thus in this situation will be produced

$${\nu _{NaH{C_2}{H_3}{O_2}}}{\text{ = 2}} \cdot {\nu _{N{a_2}C{O_3}}} = 2 \cdot 5.51[{\text{mol]}} = 11.02[{\text{mol]}}$$

moles of $NaH{C_2}{H_3}{O_2}$ .