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GenaCL600 [577]
4 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]4 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:

Number of moles of Fe = 10 mol

Number of moles of CO₂ = 15 mol

Explanation:

Given data:

Number of moles of iron oxide = 5 mol

Number of moles of carbon monoxide = 25 mol

Number of moles of product = ?

Solution:

Fe₂O₃ + 3CO   →  2Fe + 3CO₂

Now we will compare the moles of reactant with product.

                  Fe₂O₃        :         Fe

                     1             :          2

                    5             :         2×5 = 10 mol

                Fe₂O₃        :         CO₂

                     1             :          3

                    5             :         3×5 = 15 mol

                  CO           :         Fe

                     3             :          2

                    25             :         2/3×25 = 16.7 mol

                  CO            :         CO₂

                     3             :          3

                    25             :         25

Less number of moles of Fe and CO₂ are formed by iron oxide thus it will act as limiting reactant while CO is inn excess.

                 

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3 years ago
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erik [133]

Answer:

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Explanation:

The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5 because of the diagram.

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