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allochka39001 [22]

3 years ago

9

What is the percent yield of NaCl if 31g of CuCl2 reacts with excess NaNo3 to produce 21.2g of NaCl

1 answer:

Valentin [98]3 years ago

3 0

The answer is 78.7% yield.

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Phosphorous trichloride (PCl3) is produced

vladimir1956 [14]

The percentage yield obtained from the given reaction above is 74.8%

<h3>Balanced equation </h3>

P₄ + 6Cl₂ → 4PCl₃

Molar mass of P₄ = 31 × 4 = 124 g/mol

Mass of P₄ from the balanced equation = 1 × 124 = 124 g

Molar mass of PCl₃ = 31 + (35.5×3) = 137.5 g/mol

Mass of PCl₃ from the balanced equation = 4 × 137.5 = 550 g

<h3>SUMMARY</h3>

From the balanced equation above,

124 g of P₄ reacted to produce 550 g of PCl₃

<h3>How to determine the theoretical yield </h3>

From the balanced equation above,

124 g of P₄ reacted to produce 550 g of PCl₃

Therefore,

79.12 g of P₄ will react to produce = (79.12 × 550) / 124 = 350.9 g of PCl₃

<h3>How to determine the percentage yield </h3>

Actual yield of PCl₃ = 262.6 g
Theoretical yield of PCl₃ = 350.9 g

Percentage yield =?

Percentage yield = (Actual /Theoretical) × 100

Percentage yield = (262.6 / 350.9) × 100

Percentage yield = 74.8%

Learn more about stoichiometry:

brainly.com/question/14735801

4 0

2 years ago

You use 50.00 mL of a 12.0 M solution of HCl solution to make a 500.00 mL solution. What is the concentration of the new solutio

Alexandra [31]

Answer:

1.2 M

Explanation:

If you use the dilution equation (M1V1=M2V2), you end up with (50)(12)=(500)(M2), and when you solve for M2 you get 1.2 M.

6 0

3 years ago

A triangular-shaped, transparent solid is known as a ______<br><br> A. Cylinder <br> B. Prism

STatiana [176]

B.Prism because it’s triangular and a cylinder is circular

6 0

3 years ago

Read 2 more answers

I NEED HELP ASAP!!<br> I DOMT KNOW HOW TO DO THIS

KatRina [158]

What part of it are you confused about

3 0

3 years ago

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

(c) $\Delta\mu =\Delta\mu(l)-\Delta\mu(s)$ = $-S_m\DeltaT$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

6 0

2 years ago