It seems like the answer is C, you really just need to use the process of elimination.
Two types of stoichiometry are; molar mass and coefficients from balanced equation.
Answer:
13mL
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
HNO3 + KOH —> KNO3 + H2O
From the balanced equation above, we obtained the following data:
Mole ratio of the acid (nA) = 1
Mole ratio of the base (nB) = 1
Step 2:
Data obtained from the question.
This includes the following:
Molarity of the acid (Ma) = 6M
Volume of the acid (Va) =?
Volume of the base (Vb) = 39mL
Molarity of the base (Mb) = 2M
Step 3:
Determination of the volume of the acid.
Using the equation:
MaVa/MbVb = nA/nB, the volume of the acid can be obtained as follow:
MaVa/MbVb = nA/nB
6 x Va / 2 x 39 = 1/1
Cross multiply to express in linear form
6 x Va = 2 x 39
Divide both side by 6
Va = (2 x 39)/6
Va = 13mL
Therefore, the volume of the acid (HNO3) needed for the reaction is 13mL
Answer:
26.4 960 for the first one
8.9569 for the second one
