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2 years ago

7

If 30.5 g of 15% potassium nitrate solution are reacted to excess magnesium chloride how many grams of potassium chloride will b

e form

1 answer:

Nata [24]2 years ago

4 0

Here the step by step and answer

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What is the mass of silver chlorate (191.32 g/mol) that decomposes to release 0.466L of oxygen gas at STP? AgC1036) _AgCl). _026

vesna_86 [32]

Answer : The mass of silver chlorate will be 2.654 grams.

Explanation :

The balanced chemical reaction is,

$2AgClO_3\rightarrow 2AgCl+3O_2$

First we have to calculate the moles of oxygen gas at STP.

As, 22.4 L volume of oxygen gas present in 1 mole of oxygen gas

So, 0.466 L volume of oxygen gas present in $\frac{0.466}{22.4}=0.0208$ mole of oxygen gas

Now we have to calculate the moles of silver chlorate.

From the balanced chemical reaction, we conclude that

As, 3 moles of oxygen produced from 2 moles of silver chlorate

So, 0.0208 moles of oxygen produced from $\frac{2}{3}\times 0.0208=0.01387$ moles of silver chlorate

Now we have to calculate the mass of silver chlorate.

$\text{Mass of }AgClO_3=\text{Moles of }AgClO_3\times \text{Molar mass of }AgClO_3$

Molar mass of silver chlorate = 191.32 g/mole

$\text{Mass of }AgClO_3=0.01387mole\times 191.32g/mole=2.654g$

Therefore, the mass of silver chlorate will be 2.654 grams.

3 0

3 years ago

To pull up a bucket of water from a well, George pulled hard on a handle to wind up a rope. Which kind of energy was George appl

mote1985 [20]

b frictional energy

is the correct answer

4 0

3 years ago

How many formula units make up 24.2 g of magnesium chloride (MgCl2)?<br><br> Help!!

NNADVOKAT [17]

Answer:

Approximately $1.53 \times 10^{23}$ formula units ( $0.254\; \rm mol$ ).

Explanation:

Refer to a modern periodic table for the relative atomic mass of magnesium ( $\rm Mg$ ) and chlorine ( $\rm Cl$ ):

$\rm Mg$ : $24.305$ .
$\rm Cl$ : $35.45$ .

In other words, the mass of $1\; \rm mol$ of $\rm Mg$ atoms would be (approximately) $24.305\; \rm g$ .

Likewise, the mass of $1\; \rm mol$ of $\rm Cl$ atoms would be approximately $35.45\; \rm g$ .

One formula unit of the ionic compound $\rm MgCl_{2}$ includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of $1\; \rm mol$ of the formula units of this compound.

The formula $\rm MgCl_{2}$ includes one $\rm Mg$ atom and two $\rm Cl$ atoms.

Hence, every formula unit of $\rm MgCl_{2} \!$ would include the same number of atoms: one $\rm Mg\!$ atom and two $\rm Cl\!$ atoms. There would be $1\; \rm mol$ of $\rm Mg$ atoms and $2\; \rm mol$ of $\rm Cl$ atoms in $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units.

Thus, the mass of $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units would be equal to the mass of $1\; \rm mol$ of $\rm Mg$ atoms plus the mass of $2\; \rm mol$ of $\rm Cl$ atoms. (The mass of $1\; \rm mol\!\!$ of each atom could be found from the relative atomic mass of each element.)

$\begin{aligned}& M({\rm MgCl_{2}}) \\ =\; & 24.305\; {\rm g \cdot mol^{-1}} + 2\times {\rm 35.45 \; \rm g \cdot mol^{-1}} \\ =\; & 95.205\; \rm g \cdot mol^{-1}\end{aligned}$ .

In other words, the formula mass of $\rm MgCl_{2}$ is $95.205\; \rm g \cdot mol^{-1}$ .

Therefore, the number of formula units in $m = 24.2\; \rm g$ of $\rm MgCl_{2}$ would be:

$\begin{aligned}n &= \frac{m({\rm MgCl_{2}})}{M({\rm MgCl_{2}})} \\ &= \frac{24.2\; \rm g}{95.205\; \rm g\cdot mol^{-1}} \\ & \approx 0.254\; \rm mol\end{aligned}$ .

Multiple $n$ by Avogadro's Number $N_{A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}$ to estimate the number of formula units in $0.254\; \rm mol$ :

$\begin{aligned}N &= n \cdot N_{A} \\ &\approx 0.254\; \rm mol \times 6.022 \times 10^{23}\; \rm mol^{-1} \\ &\approx 1.53\times 10^{23}\end{aligned}$ .

6 0

3 years ago

How many seconds would it take to deposit 17.3 g of ag (atomic mass = 107.87) from a solution of agno3 using a current of 10.00

Brums [2.3K]

Data Given:

Time = t = ?

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 107.86/1 = 107.86 g

Amount Deposited = W = 17.3 g

Solution:
According to Faraday's Law,

W = I t e / F

Solving for t,

t = W F / I e
Putting values,
t = (17.3 g × 96500) ÷ (10 A × 107.86 g)

t = 1547.79 s

t = 1.54 × 10³ s

5 0

3 years ago

What is the mass of 3.35 mol Hg(IO3)2? 1,700 g 1,840 g 1,960 g 2,110 g

vichka [17]

Answer:- 1840 g.

Solution:- We have been given with 3.35 moles of and asked to calculate it's mass.

To convert the moles to grams we multiply the moles by the molar mass of the compound. Molar mass of the compound is the sum of atomic masses of all the atoms present in it.

molar mass of = atomic mass of Hg + 2(atomic mass of I) + 6(atomic mass of O)

= 200.59+2(126.90)+6(16.00)

= 200.59+253.80+96.00

= 550.39 gram per mol

Let's multiply the given moles by the molar mass:

$3.35mol(\frac{550.39g}{1mol})$

= 1843.8 g

Since, there are three sig figs in the given moles of compound, we need to round the calculated my to three sig figs also. So, on rounding off to three sig figs the mass becomes 1840 g.

5 0

3 years ago