NaHCO3 = 22.99 + 1.008 + 16(3) = 83.99 g/mol
Na = 22.99g/83.99 g weight of molecule =.2727 or 27.27%
3.0 g* .2727 = 0.8211 grams of sodium in sample of NaHCO3
0.8211 grams Na + 1.266 grams Cl = 2.087 grams
Answer: 26.54 grams
Explanation:
To calculate the moles :
is the limiting reagent as it limits the formation of product and 
is the excess reagent
According to stoichiometry :
As 1 moles of give = 3 moles of 
Thus 0.369 moles of 
give =
of
Mass of 
Thus 26.54 g of will be produced from the given mass.
Answer:
the initial temperature of the iron sample is Ti = 90,36 °C
Explanation:
Assuming the calorimeter has no heat loss to the surroundings:
Q w + Q iron = 0
Also when the T stops changing means an equilibrium has been reached and therefore, in that moment, the temperature of the water is the same that the iron ( final temperature of water= final temperature of iron = T )
Assuming Q= m*c*( T- Tir)
mc*cc*(T-Tc)+mir*cir*(T - Tir) = 0
Tir = 20.3 °C + 300 g * 4.186 J/g°C * (20.3 C - 19 °C) / ( 51.9 g * 0.449 J/g°C )
Tir = 90.36 °C
Note :
- The specific heat capacity of water is assumed 1 cal/g°C = 4.186 J/g°C
- We assume no reaction between iron and water
