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3 years ago

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How many grams of calcium chloride were used if 2.485 grams of calcium oxalate monohydrate were produced, assuming a 100% yield.

This means all of the calcium chloride was converted to calcium oxalate monohydrate. Report your answer to 3 decimal places in grams.

1 answer:

DochEvi [55]3 years ago

5 0

Answer:

54 G

Explanation:

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lidiya [134]

The answer is molecule

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Explanation:

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6 0

2 years ago

What volume (in liters) of a solution contains 0.14 mol of KCl?

oksano4ka [1.4K]

Answer:

$\boxed {\boxed {\sf 0.078 \ L }}$

Explanation:

We are asked to find the volume of a solution given the moles of solute and molarity.

Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:

$molarity= \frac{moles \ of \ solute}{liters \ of \ solution}$

We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.

moles of solute = 0.14 mol KCl
molarity= 1.8 mol KCl/ L
liters of solution=x

Substitute these values/variables into the formula.

$1.8 \ mol \ KCl/ L = \frac { 0.14 \ mol \ KCl}{x}$

We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

$\frac {1.8 \ mol \ KCl/L}{1} = \frac{0.14 \ mol \ KCl}{x}$

$1.8 \ mol \ KCl/ L *x = 1*0.14 \ mol \ KCl$

$1.8 \ mol \ KCl/ L *x = 0.14 \ mol \ KCl$

Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.

$\frac {1.8 \ mol \ KCl/ L *x}{1.8 \ mol \ KCl/L} = \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}$

$x= \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}$

The units of moles of potassium chloride cancel.

$x= \frac{0.14 }{1.8 L}$

$x=0.07777777778 \ L$

The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.

$x \approx 0.078 \ L$

There are approximately <u>0.078 liters of solution.</u>

5 0

2 years ago

Excess magnesium reacts with 165.0 grams of hydrochloric acid in a single displacement reaction.

JulsSmile [24]

Answer:

The volume of hydrogen gas produced will be approximately 50.7 liters under STP.

Explanation:

Relative atomic mass data from a modern periodic table:

H: 1.008;
Cl: 35.45.

Magnesium is a reactive metal. It reacts with hydrochloric acid to produce

Hydrogen gas $\rm H_2$ , and
Magnesium chloride, which is a salt.

The chemical equation will be something like

$\rm ?\;Mg\;(s) + ?\;HCl \;(aq)\to ?\;H_2 \;(g)+ [\text{Formula of the Salt}]$ ,

where the coefficients and the formula of the salt are to be found.

To determine the number of moles of $\rm H_2$ that will be produced, first find the formula of the salt, magnesium chloride.

Magnesium is a group 2 metal. The oxidation state of magnesium in compounds tends to be +2.

On the other hand, the charge on each chloride ion is -1. Each magnesium ion needs to pair up with two chloride ions for the charge to balance in the salt, magnesium chloride. The formula for the salt will be $\rm MgCl_2$ .

$\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ ?\;MgCl_2\;(aq)$ .

Balance the equation. $\rm MgCl_2$ contains the largest number of atoms among all species in this reaction. Start by setting its coefficient to 1.

$\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ {\bf 1\;MgCl_2}\;(aq)$ .

The number of $\rm Mg$ and $\rm Cl$ atoms shall be the same on both sides. Therefore

$\rm {\bf 1\;Mg}\;(s) + {\bf 2\;HCl}\;(aq) \to ?\;H_2 \;(g)+ {1\;\underset{\wedge}{Mg}\underset{\wedge}{Cl_2}}\;(aq)$ .

The number of $\rm H$ atoms shall also conserve. Hence the equation:

$\rm {1\;Mg}\;(s) + {2\;\underset{\wedge}{H}Cl}\;(aq) \to {\bf 1\;H_2 \;(g)}+ {1\;MgCl_2}\;(aq)$ .

How many moles of HCl are available?

$M(\rm HCl) = 1.008 + 35.45 = 36.458\;g\cdot mol^{-1}$ .

$\displaystyle n({\rm HCl}) = \frac{m(\text{HCl})}{M(\text{HCl})} = \rm \frac{165.0\;g}{36.458\;g\cdot mol^{-1}} = 4.52576\;mol$ .

How many moles of Hydrogen gas will be produced?

Refer to the balanced chemical equation, the coefficient in front of $\rm HCl$ is 2 while the coefficient in front of $\rm H_2$ is 1. In other words, it will take two moles of $\rm HCl$ to produce one mole of $\rm H_2$ . $\rm 4.52576\;mol$ of $\rm HCl$ will produce only one half as much $\rm H_2$ .

Alternatively, consider the ratio between the coefficient in front of $\rm H_2$ and $\rm HCl$ is:

$\displaystyle \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}$ .

$\displaystyle n(\text{H}_2) = n(\text{HCl})\cdot \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}\;n(\text{HCl}) = \rm \frac{1}{2}\times 4.52576\;mol = 2.26288\;mol$ .

What will be the volume of that many hydrogen gas?

One mole of an ideal gas occupies a volume of 22.4 liters under STP (where the pressure is 1 atm.) On certain textbook where STP is defined as $\rm 1.00\times 10^{5}\;Pa$ , that volume will be 22.7 liters.

$V(\text{H}_2) = \rm 2.26288\;mol\times 22.4\;L\cdot mol^{-1} = 50.69\; L$ , or

$V(\text{H}_2) = \rm 2.26288\;mol\times 22.7\;L\cdot mol^{-1} = 51.37\; L$ .

The value "165.0 grams" from the question comes with four significant figures. Keep more significant figures than that in calculations. Round the final result to four significant figures.

5 0

3 years ago

5. What is the partial pressure of the remaining gas, if the total is 384 torr 2

alexgriva [62]

Answer: 83.11 torr

Explanation:

According to Dalton's Law of partial pressure, the total pressure of a mixture of gases is the sum of the pressure of each individual gas.

i.e Ptotal = P1 + P2 + P3 + .......

In this case,

Ptotal = 384 torr

P1 = 289 torr

P2 = 11.89 torr

P3 = ? (let the partial pressure of the remaining gas be Z)

Ptotal = P1 + P2 + Z

384 torr = 289 torr + 11.89 torr + Z

384 torr = 300.89 torr + Z

Z = 384 torr - 300.89 torr

Z = 83.11 torr

Thus, the partial pressure of the remaining gas is 83.11 torr.

5 0

3 years ago