Ask question

Ask question

All categories

3 years ago

9

I begin the reaction with 0.45 g of beryllium. If my actual yield of beryllium chloride (mm = 79.91 g/mol) was 3.5 grams, what w

as my percent yield?
Please show work

1 answer:

trasher [3.6K]3 years ago

5 0

The percentage of yield was 777.78%

<u>Explanation:</u>

We have the equation,

Be [s] + 2 HCl [aq] → BeCl 2(aq] + H 2(g] ↑ Be (s] + 2 HCl [aq] → BeCl 2(aq] + H 2(g] ↑

To find the percent yield we have the formula

Percentage of Yield= what you actually get/ what you should theoretically get x 100

=3.5 g/0.45 g 100

= 777.78 %

The percentage of yield was 777.78%

You might be interested in

Between Lab Period 1 and Lab Period 2, design a separation scheme for all 4 cations. Use the results of your preliminary tests a

fiasKO [112]

Answer:

SEPARATION SCHEME FOR CATIONS

GIVEN CATIONS : $Ag^{+} \ , Fe^{3+} , Cu^{2+}, Ni^{2+}$

Step 1: Add $6mol/dm^3$ of $HCl$ to the mixture solution

Result : This would cause a precipitate of $AgCl$ to be formed

Reaction : $Ag^{+} _{(aq)} + Cl^{-} _{(aq)} ---------> AgCl(ppt)$

Step 2 : Next is to remove the precipitate and add $H_2S$ to the remaining

solution in the presence of $0.2 \ mol/dm^3$ of HCl

Result : This would cause a precipitate of $CuS$ to be formed

Reaction : $Cu^{2+}_{(aq)} + S^{2-}_{(aq)} ------> Cu_2S(ppt)$

Step 3: Next remove the precipitate then add $6 \ mol/dm^3$ of aqueous

$NH_3 (NH_3 \cdot H_2 O)$ , process the solution in a centrifuge,when the

process is done then sort out the precipitate from the solution

Now this precipitate is $Fe(OH)_3$ and the remaining solution

contains $(Ni (NH_3)_6)$

Next take out the precipitate to a different beaker and add HCl

to it this will dissolve it, then add a drop of $NH_4SCN$ this will

form a precipitate $Fe(SCN)_{6}^{3-}$ which will have the color of

blood indicating the presence of $Fe^{3+}$

Reaction : $F^{3+}_{(aq)} + 30H^-_{(aq)} --------->Fe(OH)_3_{(aq)}$

$Fe (OH)_{(s)} _3 + 3H^{+}_{aq} -------> Fe^{3+}_{aq} + 3H_2O_{(l)}$

$Fe^{3+} + 6SCN^{-} -----> Fe(SCN)_6 ^{3-}$

Now the remaining mixture contains $Ni^{2+}$

Explanation:

6 0

3 years ago

Given the equation C3H8(g) + O2(g) = CO2(g) + H2O(g) and that the enthalpies of formation for H2O(g) = -241.8 kJ/mol, CO2(g) = -

kari74 [83]

Answer: 72.4 kJ/mol

Explanation:

The balanced chemical reaction is,

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)$ $\Delta H=-2220.1kJ/mol$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})]$

where,

n = number of moles

$\Delta H_{O_2}=0$ (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

$-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})]$

$\Delta H_{C_3H_8}=72.4kJ/mol$

Therefore, the heat of formation of propane is 72.4 kJ/mol

7 0

2 years ago

Bezzdna [24]

Answer:

19.5g is the theoretical yield of alum

Explanation:

Based on the balanced reaction, 4 moles of sulfuric acid produce 2 moles of alum. To solve this question we need to find the moles of H2SO4. With these moles we can find the moles of alum and its mass assuming all sulfuric acid reacts producing alum.

<em>Moles Sulfuric Acid:</em>

8.3mL = 0.0083L * (9.9mol/L) = 0.08217 moles sulfuric acid

<em>Moles Alum:</em>

0.08217 moles sulfuric acid * (2mol KAl(SO4)2•12H2O / 4mol H2SO4) =

0.041085 moles KAl(SO4)2•12H2O

<em>Mass Alum -Molar mass: 474.3884 g/mol-</em>

0.041085 moles KAl(SO4)2•12H2O * (474.3884 g/mol) =

<h3>19.5g is the theoretical yield of alum</h3>

3 0

2 years ago

⚠️⚠️Plz Help⚠️⚠️

Dmitrij [34]

Answer:

B

Explanation:

just took the test

6 0

2 years ago

A formula that relates the amount of a substance to a mass?

vagabundo [1.1K]

I believe its n= m/M

n - number of moles
m - mass of substance
M - molar mass of substance

7 0

2 years ago

Read 2 more answers