Question

Can someone check this for me?
The lab results were Evaporating Dish 24.35g
NaCHO3 37.06 (includes evaporating dish)
HCI 40.06g (includes evaporating dish)

The question is : 3. The NaHCO3 is the limiting reactant and the HCl is the excess reactant in this experiment. Determine the theoretical yield of the NaCl product, showing all of your work in the space below.

My answer is :
NaCHO3: 37.06g - 24.35g = 12.71 g NaHCO3
1 mol NaHCO3 & 1 HCl --> 1 mol NaCl & 1 H2O & 1 CO2
12.71 g NaHCO3 @ 58.44 g/mol NaCl / 84.01 g/mol NaHCO3 = 8.841 grams of NaCl

Answer 1

\(\huge mole= \frac{ mass}{ Molecular ~mass } \)

\(\huge mass= mole\times molecular~ mass\)

mass= m
Molecular mass= Mr

\(\huge m= \frac{m ~of~NaHCO_3}{Mr~of~NaHCO_3}\times Mr~ of~ NaCl \)

thats what you did :)
so yeah thats right
since the ration as you wrote is 1:1

Answer 2

ratio*

Answer 3

Thank you!!!