Question

Part 1 (1 pt) See Periodic Table D See Hint Balance this reaction: x x. 'He , o, 2 HNO3 + Na,coz — 2 NaNO3 + CO2 + 420 Part 2 (1 pt) D See Hint If you have 520.0 g of HNO3 (63.01 g/mol) and excess Na2CO3, how many grams of carbon dioxide (CO2, 44.01 g/mol) will be produced?

Answer 1

1)

The reaction is:

HNO3 + Na2CO3 -> NaNO3 + CO2 + H2O

Balance H:

2 HNO3 + Na2CO3 -> NaNO3 + CO2 + H2O

Balance Na:

2 HNO3 + Na2CO3 -> 2 NaNO3 + CO2 + H2O

The equation is now balanced

Answer:

2 HNO3 + Na2CO3 -> 2 NaNO3 + CO2 + H2O

2)

Molar mass of HNO3 = 63.01 g/mol

mass of HNO3 = 5.2*10^2 g

mol of HNO3 = (mass)/(molar mass)

= 5.2*10^2/63.01

= 8.252 mol

According to balanced equation

mol of CO2 formed = (1/2)* moles of HNO3

= (1/2)*8.252

= 4.126 mol

Molar mass of CO2 = 44.01 g/mol

mass of CO2 = number of mol * molar mass

= 4.126*44.01

= 1.82 g

Answer: 182 g