What volume (in mL) of a 0.300 M HNO3 solution will completely react with 43.9 mL of a 0.146 M Na2CO3 solution according to this balanced chemical equation?
Na2CO3(aq)+2 HNO3(aq)->2 NaNO3(aq)+CO2(g)+H20(l)
In the balanced chemical equation, the stoichiometric ratio between Na2CO3 and HNO3 is 1:2. This means that 1 mole of Na2CO3 reacts with 2 moles of HNO3.
First, let's calculate the number of moles of Na2CO3 in the given volume of the Na2CO3 solution:
moles of Na2CO3 = concentration of Na2CO3 × volume of Na2CO3 solution
= 0.146 M × 43.9 mL
= 0.0064034 moles
According to the stoichiometry of the balanced equation, since the ratio of Na2CO3 to HNO3 is 1:2, the number of moles of HNO3 required will be twice the number of moles of Na2CO3:
moles of HNO3 required = 2 × moles of Na2CO3
= 2 × 0.0064034 moles
= 0.0128068 moles
Now, we can calculate the volume of the HNO3 solution needed to react completely with the Na2CO3 solution:
volume of HNO3 solution = moles of HNO3 required / concentration of HNO3
Given that the concentration of HNO3 is 0.300 M:
volume of HNO3 solution = 0.0128068 moles / 0.300 M
= 0.0427 L
= 42.7 mL
Therefore, 42.7 mL of the 0.300 M HNO3 solution will completely react with 43.9 mL of the 0.146 M Na2CO3 solution.
To find the volume of the HNO3 solution that will completely react with the Na2CO3 solution, you can use the concept of stoichiometry.
First, calculate the number of moles of Na2CO3 using its molarity and volume:
moles of Na2CO3 = molarity × volume
= 0.146 M × 43.9 mL
= 6.4054 × 10^(-3) mol
When balancing the chemical equation, we see that the ratio between Na2CO3 and HNO3 is 1:2. Hence, the number of moles of HNO3 required will be twice the number of moles of Na2CO3:
moles of HNO3 = 2 × moles of Na2CO3
= 2 × 6.4054 × 10^(-3) mol
= 1.28108 × 10^(-2) mol
Now, we need to find the volume of the HNO3 solution that corresponds to this number of moles. We can do this by dividing the number of moles by the molarity of HNO3:
volume of HNO3 = moles of HNO3 / molarity
= 1.28108 × 10^(-2) mol / 0.300 M
= 4.270 m
Therefore, the volume of the 0.300 M HNO3 solution required to completely react with 43.9 mL of the 0.146 M Na2CO3 solution is 4.270 mL.