Question

(Given Sodium Carbonate: 15.900g)

Molar Mass = 105.989g/mol

1. Calculate the mass of sodium chloride that is expected to be produced. (molar mass sodium chloride = 58.440g/mol)

2. Calculate the mass of water expected to be produced. (molar mass of water = 18.0153g/mol)

3. Calculate the mass of carbon dioxide expected to be produced. (molar mass of carbon dioxide = 44.0095g/mol)

Please print/type equations/steps clearly.

Answer 1

Answer:-

(1)-

Given:-

mass of sodium Carbonate (Na₂CO₃) = 15.900 g

Molar Mass of sodium Carbonate (Na₂CO₃) = 105.989 g / mol

molar mass of sodium chloride (NaCl) = 58.440 g / mol

mass of sodium chloride (NaCl) that is expected to be produced = ?

As we know that when sodium Carbonate (Na₂CO₃) reacts with HCl to produced the sodium chloride(NaCl) as follows:-

Na₂CO₃ + 2HCl $\rightarrow$    2NaCl + CO₂ + H₂O

105.989 g 2 $\times$ 36.460 g 2 $\times$ 58.440 g   44.0095 g   18.0153g

105.989 g 72.920 g 116.88 g 44.0095 g   18.0153 g

As mentioned above

105.989 g of sodium Carbonate (Na₂CO₃) produced = 116.88 g of sodium chloride(NaCl)

1.0 g of sodium Carbonate (Na₂CO₃) produced = 116.88 / 105.989 g of sodium chloride(NaCl)

then

15.900 g of sodium Carbonate (Na₂CO₃) produced = 116.88 $\times$ 15.900 / 105.989 g of sodium chloride(NaCl)

15.900 g of sodium Carbonate (Na₂CO₃) produced = 1858.392‬  / 105.989 g of sodium chloride(NaCl)

15.900 g of sodium Carbonate (Na₂CO₃) produced = 17.534 g of sodium chloride(NaCl)

therefore

mass of sodium chloride (NaCl) that is expected to be produced = 17.534 g  (i.e the answer)

(2)-

Given:-

mass of sodium Carbonate (Na₂CO₃) = 15.900 g

Molar Mass of sodium Carbonate (Na₂CO₃) = 105.989 g / mol

molar mass of water (H₂O) = 18.0153 g / mol

mass of water (H₂O) expected to be produced = ?

As we know that when sodium Carbonate (Na₂CO₃) reacts with HCl to produced the water (H₂O) as follows:-

Na₂CO₃ + 2HCl $\rightarrow$    2NaCl + CO₂ + H₂O

105.989 g 2 $\times$ 36.460 g 2 $\times$ 58.440 g   44.0095 g   18.0153g

105.989 g 72.920 g 116.88 g 44.0095 g   18.0153 g

As mentioned above

105.989 g of sodium Carbonate (Na₂CO₃) produced = 18.0153 g water (H₂O)

1.0 g of sodium Carbonate (Na₂CO₃) produced = 18.0153 / 105.989 g water (H₂O)

then

15.900 g of sodium Carbonate (Na₂CO₃) produced = 18.0153 $\times$ ​​​​​​​ 15.900 / 105.989 g water (H₂O)

15.900 g of sodium Carbonate (Na₂CO₃) produced = 286.44327‬‬  / 105.989 g water (H₂O)

15.900 g of sodium Carbonate (Na₂CO₃) produced = 2.7026 g water (H₂O)

therefore

mass of water (H₂O) expected to be produced =  2.7026 g  (i.e the answer).

(3)-

Given:-

mass of sodium Carbonate (Na₂CO₃) = 15.900 g

Molar Mass of sodium Carbonate (Na₂CO₃) = 105.989 g / mol

molar mass of carbon dioxide (CO₂) = 44.0095 g / mol

mass of carbon dioxide (CO₂) expected to be produced = ?

As we know that when sodium Carbonate (Na₂CO₃) reacts with HCl to produced the carbon dioxide (CO₂) as follows:-

Na₂CO₃ + 2HCl $\rightarrow$    2NaCl + CO₂ + H₂O

105.989 g 2 $\times$ 36.460 g 2 $\times$ 58.440 g   44.0095 g 18.0153g

105.989 g 72.920 g 116.88 g 44.0095 g 18.0153 g

As mentioned above

105.989 g of sodium Carbonate (Na₂CO₃) produced = 44.0095 g carbon dioxide (CO₂)

1.0 g of sodium Carbonate (Na₂CO₃) produced = 44.0095 / 105.989 g carbon dioxide (CO₂)

then

15.900 g of sodium Carbonate (Na₂CO₃) produced = 44.0095 $\times$ ​​​​​​​ 15.900 / 105.989 g carbon dioxide (CO₂)

15.900 g of sodium Carbonate (Na₂CO₃) produced = 699.75105‬  / 105.989 g carbon dioxide (CO₂)

15.900 g of sodium Carbonate (Na₂CO₃) produced = 6.6021 g carbon dioxide (CO₂)

therefore

  mass of carbon dioxide (CO₂) expected to be produced =  6.6021 g  (i.e the answer).