How many grams of calcium chloride are needed to produce 1.50 g of potassium chloride? cacl2(aq) + k2co3(aq) → 2 kcl(aq) + caco3
(aq) 1.12 g 2.23 g 0.896 g 4.47 g?
2 answers:
The calcium chloride needed to produce 1.5 grams of KCl is: 1.12 grams
<h3><em>Further explanation</em></h3>Stoichiometry in Chemistry learn about chemicals mainly emphasizes quantitative, such as the calculation of volume, mass, number, which is related to numbers, molecules, elements, etc.
A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.
In the reaction there are also manifestations of reagent substances namely gas (g), liquid (liquid / l), solid (solid / s) and solution (aqueous / aq).
Reactions that occur:
CaCl₂ (aq) + K₂CO₃ (aq) → 2 KCl (aq) + CaCO₃ (aq)
From the reaction equation above, the reaction coefficient shows the mole ratio of reagents and products
mol CaCl₂: moles of KCl = 1: 2
relative molecular mass KCl = 74.6
relative molecular mass of CaCl2 = 111
then the mole:
mole KCl = mass: relative molecular mass
mole KCl = 1.5 gram: 74.6
mole KCl = 0.020107
so that
mole CaCl₂ = 1/2. 0.020107
mole CaCl₂ = 0.0100535
So the mass of CaCl2 = mole x relative molecular mass
mass CaCl₂ = 0.0100535 x 111
mass CaCl₂ = 1.12 gram
<h3><em>Learn more</em></h3>moles of water
grams of oxygen required
mass of copper required
Keywords: CaCl2, KCl, mole, mass, relative molecular mass
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Answer:
- Please, find the graph with the labels and points located on the axes in the picture attached.
Explanation:
This is how you meet all the instructions and some important comments to understand how this kind of graphs word:
<u>1) Label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL. </u>
The horizontal axis is used to record the independent variable and the vertical axis is used to record the dependent variable. The axes most be properly labeled with the name of the variable and the units.
In this case the origin is the point (0,0) which means that the axes cross each other, perpendicularly, at a pressure of 0 mb and a volume of 0.0 mililiters.
<u>2) Assign values to axes divisions in such a way that you occupy almost all the space on both axes. </u>
A good graph searches to occupy the whole space on both cases; to do that, find the maximum value for each variable, pressure and volume, and choose the values of the marks.
The range of the pressure (horizontal axis) is [90, 760 mb], so you should choose big divisions (marks) of 100 mb, and assign 800 mb to the right most mark on the horizontal axis. Then, you can divide each interval of 100 mb into 10 spaces, with small divisions of 10 mb (my graph uses 4 spcaes, with small divisions of 25 mb, but I recommend you use small divisions of 10 mb).
The range of the volume (vertical axis) is [0.1, 0.8], so you should choose only divisions with value of 0.1 ml.
<u>3) Now locate and label the points: </u>
- (90, 0.9) ⇒ 90 mb, 0.9 ml
- (100, 0.8) ⇒ 100 mb, 0.8 ml
- (400, 0.2) ⇒ 400 mb, 0.2 ml
- (600, 0.15) ⇒ 600 mb, 0.15 ml
- (760, 0.1) ⇒ 760 mb, 0.1 ml
The points of the kind (x, y) are called ordered pairs, which means that the order matters, because it has a meaning: the first number represents the independent variable and the second number represents the dependent variable.
So, in the point (90, 0.9), 90 is a pressure of 90 mb and 0.9 is a volume of 0.9 ml.
To locate (600, 0.15), since the horizontal marks have value of 0.1, you must locate the second coordinate of your point between the marks 0.1 and 0.2 ml.
With that you can now locate each point on your graph.
