Question

2. How many moles of HCI are needed to react with 0.47 g of NaHC03 ?

Answer 1

HCl reacts with NaHCO₃ this way:

$HCl_{}+NaHCO_3\rightarrow NaCl+H_2O+CO_2$

The equation is already balanced, it means that 1 mole of HCl reacts with 1 mole of NaHCO₃.

Convert the mass of NaHCO₃ to moles using the molecular mass of NaHCO₃ which is 84g/mol.

$0.47g\cdot\frac{1mol}{84g}=0.0056mol$

Now, use the ration stated before to find how many moles of HCl react with this amount of NaHCO₃:

$0.0056mol\cdot\frac{1mol}{1mol}=0.0056mol$

It means that 0.0056 moles of HCl are needed to react with 0.47g of NaHCO₃.