Question

A****41.9 mL of a 0.146 M Na2CO3 solution completely react with a 0.150 MHNO3 solution according to the following balanced chemical equation:  

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

What mass (in grams) of carbon dioxide is formed?

B*****

What volume (in mL) of a 0.100 MHNO3 solution is required to completely react with 31.8 mL of a 0.108 MNa2CO3 solution according to the following balanced chemical equation?

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

C****

What volume of a 5.70 M solution of NaNO3 do you need to make 0.510 L of a 1.00 M solution of NaNO3?

if you could write out the steps that would be so helpful! thank you

Answer 1

The reaction is

Na₂CO₃(aq)+2HNO₃(aq)→2NaNO₃(aq)+CO₂(g)+H₂O(l)

A) 41.9 ml 0.146M Na₂CO₃ contain

41.9 ml = 0.0419 dm³

Number of mole of solute = Molarity Volume of solution in dm³

= 0.146 0.0419 = 0.006117 mole

41.9 ml 0.146M Na₂CO₃ contain 0.006117 mole.

According to reaction 1 mole of Na₂CO₃ produce 1 mole of CO₂ then 0.06117 mole of Na₂CO₃ produce 0.006117 mole of CO₂

1 mole of CO₂ =44.01g/mole then 0.006117 mole = 44.010.06117/1 =0.2692 gm

0.2692 gm of CO₂ formed.

B)31.8 ml 0.108 M Na₂CO₃=

31.8 ml = 0.0318 dm³

Number of mole of solute = Molarity Volume of solution in dm³

= 0.1080.0318 = 0.0034344 mole

According to reaction 1 mole of Na₂CO₃ react with 2 mole of HNO₃ then 0.034344 mole of Na₂CO₃ react with 0.00343442 =0.0068688 mole of HNO₃

Volume of solution in dm³ = Number of mole of solute /Molarity

= 0.0068688/0.100 = 0.068688 dm³

0.68688 dm³ = 68.688 ml

68.688 ml of a 0.100 MHNO3 solution is required to completely react with 31.8 mL of a 0.108 MNa2CO3 solution .

C) C₁V₁ = C₂V₂

V₁= C₂V₂/C₁

= 1.00M510 ml/5.70M = 89.47 ml

89.47 ml of a 5.70 M solution of NaNO3 need to make 0.510 L of a 1.00 M solution of NaNO3.