Answer:
6.07 g.
Step-by-step explanation:
Hello!
In this case, given the chemical reaction:
![Zn + H_2CO_3 \rightarrow ZnCO_3 + H_2](https://img.qammunity.org/2022/formulas/chemistry/college/d8fsbr2dg6s6thszdsvm99v6c83ox59zho.png)
We first need to compute the moles of each reactant considering their molar masses:
![n_(Zn)=4gZn*(1molZn)/(65.38 gZn) =0.0612molZn\\\\n_(H_2CO_3)=3gH_2CO_3*(1molH_2CO_3)/(62.03 gH_2CO_3) =0.0484molH_2CO_3](https://img.qammunity.org/2022/formulas/chemistry/college/l7en7kyq6boptg4bnchgyw77rq0g0j614y.png)
Now, compute the mass of zinc carbonate yielded by each reactant in order to pick out the correct limiting reactant:
![n_(ZnCO3)^(by\ Zn) = 0.0612molZn*(1molZnCO3)/(1molZn)*(125.4 gZnCO3)/(1molZnCO3)=7.67gZnCO3 \\\\n_(ZnCO3)^(by\ H_2CO3) = 0.0484molH_2CO3*(1molZnCO3)/(1molH_2CO3)*(125.4 gZnCO3)/(1molZnCO3)=6.07gZnCO3](https://img.qammunity.org/2022/formulas/chemistry/college/ypms2vfrpszkbol793gtxp8u34ng0303qr.png)
Thus, we conclude carbonic acid is the limiting reactant as it produces the fewest grams of product and therefore the yielded mass of zinc carbonate product 6.07 g.
Best regards.