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GenaCL600 [577]
3 years ago
5

How many grams of calcium nitrate are needed to make 3.30 L of a 0.10 M solution?

Chemistry
1 answer:
ki77a [65]3 years ago
6 0
54.15 g
First you start out with the equation n=cv (n= moles, c=molarity, v= volume)
You’re going to multiply 0.10M by 3.30L to get an answer of 0.33 moles of Ca(NO3)2
From there you’re gonna convert the moles to grams to get your answer, first you have to find the molar mass of Ca(NO3)2
This can be done by finding adding the molar mass of each individual substance
The answer you should get for the molar mass is 164.1 g
From there just multiply the number of moles you calculated (0.33 mol) by the molar mass (164.1 g) and your answer is going to be 54.15 g Ca(NO3)2
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Answer:

Option D. 30 mL.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HNO3 + KOH —> KNO3 + H2O

From the balanced equation above,

The mole ratio of the acid, nA = 1

The mole ratio of the base, nB = 1

Step 2:

Data obtained from the question. This include the following:

Volume of base, KOH (Vb) =.?

Molarity of base, KOH (Mb) = 0.5M

Volume of acid, HNO3 (Va) = 10mL

Molarity of acid, HNO3 (Ma) = 1.5M

Step 3:

Determination of the volume of the base, KOH needed for the reaction. This can be obtained as follow:

MaVa / MbVb = nA/nB

1.5 x 10 / 0.5 x Vb = 1

Cross multiply

0.5 x Vb = 1.5 x 10

Divide both side by 0.5

Vb = (1.5 x 10) /0.5

Vb = 30mL

Therefore, the volume of the base, KOH needed for the reaction is 30mL.

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