Question

For the reaction shown, calculate how many grams of each product form when the following amounts of reactant completely react to form products. Assume that there is more than enough of the other reactant. 2HCl(aq)+Na2CO3(aq)→2NaCl(aq)+H2O(l)+CO2(g)

10.6 gHCl

a. mNaCl=

b. mH2O =

c. mCO2 =

10.6 gNa2CO3

a. mNaCl=

b. mH2O =

c. mCO2 =

Answer 1

1)

a)

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass of HCl = 10.6 g

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

= 10.6/36.46

= 0.2907 mol

According to balanced equation

mol of NaCl formed = moles of HCl

= 0.2907 mol

Molar mass of NaCl,

MM = 1*MM(Na) + 1*MM(Cl)

= 1*22.99 + 1*35.45

= 58.44 g/mol

mass of NaCl = number of mol * molar mass

= 0.2907*58.44

= 16.99 g

Answer: 17.0 g

b)

According to balanced equation

mol of H2O formed = (1/2)* moles of HCl

= (1/2)*0.2907

= 0.1454 mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass of H2O = number of mol * molar mass

= 0.1454*18.02

= 2.619 g

Answer: 2.62 g

C)

According to balanced equation

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

= (1/2)*0.2907

= 0.1454 mol

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass of CO2 = number of mol * molar mass

= 0.1454*44.01

= 6.398 g

Answer: 6.40 g

2)

a)

Molar mass of Na2CO3,

MM = 2*MM(Na) + 1*MM(C) + 3*MM(O)

= 2*22.99 + 1*12.01 + 3*16.0

= 105.99 g/mol

mass of Na2CO3 = 10.6 g

mol of Na2CO3 = (mass)/(molar mass)

= 10.6/1.06*10^2

= 0.1 mol

According to balanced equation

mol of NaCl formed = (2/1)* moles of Na2CO3

= (2/1)*0.1

= 0.2 mol

Molar mass of NaCl,

MM = 1*MM(Na) + 1*MM(Cl)

= 1*22.99 + 1*35.45

= 58.44 g/mol

mass of NaCl = number of mol * molar mass

= 0.2*58.44

= 11.69 g

Answer: 11.7 g

b)

According to balanced equation

mol of H2O formed = moles of Na2CO3

= 0.1 mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass of H2O = number of mol * molar mass

= 0.1*18.02

= 1.802 g

Answer: 1.80 g

c)

According to balanced equation

mol of CO2 formed = moles of Na2CO3

= 0.1 mol

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass of CO2 = number of mol * molar mass

= 0.1*44.01

= 4.401 g

Answer: 4.40 g