What is the mole ratio of the following equation?

The compound cisplatin, pt(nh3)2cl2 , has been studied extensively as an antitumor agent.

Nataliya [291]

a. Elemental percent composition is the mass percent of each element in the compound.

The formula for mass elemental percent composition = $\frac{mass of element}{mass of compound}$ (1)

The molecular formula of cisplatin is $Pt(NH_3)_2Cl_2$ .

The atomic weight of the elements in cisplatin is:

Platinum, $Pt = 195.084 u$

Nitrogen, $N = 14.0067 u$

Hydrogen, $H = 1 u$

Chlorine, $Cl = 35.453 u$

The molar mass of $Pt(NH_3)_2Cl_2$ = $195.084+ (2\times 14.0067)+(6\times 1)+(2\times 35.453)$ = $300.00 g/mol$

The mass of each element calculated using formula (1):

- Platinum, $Pt$ %

$\frac{195.084}{300.00} \times 100 = 65.23$ %.

- Nitrogen, $N$ %

$\frac{2\times 14.0067}{300.00} \times 100 = 9.34$ %

- Hydrogen, $H$ %

$\frac{6\times 1}{300.00} \times 100 = 2.0$ %

- Chlorine, $Cl$ %

$\frac{2\times 35.453}{300.00} \times 100 = 23.63$ %

b. The given reaction of cisplatin is:

$K_2PtCl_4(aq)+2NH_3(aq)\rightarrow Pt(NH_3)_2Cl_2(s)+2KCl(aq)$

According to the balanced reaction, 1 mole of $K_2PtCl_4$ gives 1 mole of $Pt(NH_3)_2Cl_2$ .

Now, calculating the number of moles of $K_2PtCl_4$ in 100.0 g.

Number of moles = $\frac{given mass}{Molar mass}$

Molar mass of $K_2PtCl_4$ = $2\times 39.0983+195.084+4\times 35.453 = 415.093 g/mol$

Number of moles of $K_2PtCl_4$ = $\frac{100 g}{415.093 g/mol} = 0.241 mole$ .

Since, 1 mole of $K_2PtCl_4$ gives 1 mole of $Pt(NH_3)_2Cl_2$ . Therefore, mass of cisplatin is:

$0.241 mole\times 300 g/mol = 72.3 g$

For mass of $KCl$ :

Molar mass of $KCl$ = $39.0983 + 35.453 = 74.55 g/mol$

Since, 1 mole of $K_2PtCl_4$ gives 2 mole of $KCl$ . Therefore, mass of $KCl$ is:

$0.241 mole\times 74.55 g/mol\times 2 = 35.93 g$

5 0

3 years ago

What is the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm?

Volgvan

Answer: The new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

Explanation:

Given: $V_{1}$ = 61 L, $T_{1}$ = 183 K, $P_{1}$ = 0.60 atm

At STP, the value of pressure is 1 atm and temperature is 273.15 K.

Now, formula used to calculate the new volume is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{0.60 atm \times 61 L}{183 K} = \frac{1 atm \times V_{2}}{273.15 K}\\V_{2} = 54.63 L$

Thus, we can conclude that the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

4 0

2 years ago

A 2.5% (by mass) solution concentration signifies that there is of solute in every 100 g of solution. 2. therefore, when 2.5% is

densk [106]

Answer #1 is "there is 2.5 grams of solute in every 100 g of solution."
We calculate for 2.5% by mass solution by dividing the mass of the solute by the mass of the solution and then multiply by 100.
Answer #2 is "that mass ratio would be 2.5/100 or 2.5 grams of solute/100 grams of solution."
We weigh out 2.5 grams of solute and then add 97.5 grams of solvent to make a total of 100 gram solution, that is,
mass of solute / mass of solution = 2.5g solute / (2.5g solute + 97.5g solvent)
= 2.5g solute / 100g solution
Answer#3 is "a solution mass of 1 kg is 10 times greater than 100 g, thus one kilogram (1 kg) of a 2.5% ki solution would contain 25 grams of ki."
We multiply 10 to each mass so that 100 grams becomes 1000grams since 1000 grams is equal to 1 kg:
mass of solute / mass of solution = 2.5g*10/[(2.5g*10) + (97.5g*10)]
= 25g solute/(25g solute + 975g solvent)
= 25g solute/1000g solution
= 25g solute/1kg solution

5 0

3 years ago

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What is the molecular formula of a compound containing 89% cesium (Cs) and 11% oxygen (O) with a molar mass = 298 g/mol?

Maurinko [17]

M(Cs)=133 g/mol
M(O)=16 g/mol
M(CsxOy)=298 g/mol
w(Cs)=0.89
w(O)=0.11

CsxOy

x=M(CsxOy)w(Cs)/M(Cs)
x=298*0.89/133=2

y=M(CsxOy)w(O)/M(O)
y=298*0.11/16=2

Cs₂O₂ cesium peroxide

6 0

3 years ago

Read 2 more answers

Please answer quick

JulijaS [17]

The paper will turn red

6 0

3 years ago