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2 years ago

CuSO4(aq) + 2NaOH(aq) Cu(OH)2(s) + Na2SO4(aq)

1 answer:

6 0

Your answer is right.

Important elements to consider:

- to use the balanced equation (which you did)
- divide the masses of each compound by the correspondant molar masses (which you did)
- compare the theoretical proportions with the current proportions

Theoretical: 2 mol of Na OH : 1 mol of CuSO4
Then 4 mol of NaOH need 2 mol of CUSO4.

Given that you have more than 2 mol of of CUSO4 you have plenty of it and the NaOH will consume first, being this the limiting reagent.

You might be interested in

choose the reaction that illustrates delta H *f for Ca(NO3)2.(A) Ca (s) + N2 (g) + 3O2 ---> Ca(NO3)2 (s)(B) Ca2 (aq) + 2 NO3-

kompoz [17]

Answer: The correct answer is Option A.

Explanation:

Standard enthalpy of formation is the change in enthalpy of one mole of a substance present at the standard state that is 1 atm of pressure and 298 K of temperature. The substance is formed from its pure elements under the same conditions.

We are given a chemical compound having chemical formula $Ca(NO_3)_2$

This compound is formed by the combination of calcium, nitrogen and oxygen elements.

The chemical equation for the formation of $Ca(NO_3)_2$ from the components in their standard states follows:

$Ca+N_2+3O_2\rightarrow Ca(NO_3)_3$

Hence, the correct answer is Option A.

3 0

2 years ago

If you were to assume a 95% yield for the formation of the phosphonium ion, how many milligrams of benzyl chloride and triphenyl

luda_lava [24]

I am attempting the problem for phosphonium Ion rather than its chloride salt. The chemical equation is shown below along with molar masses in mg.

First of all we will calculate the amounts of reactants required for the synthesis of 220 mg of phophonium ion. Calculations for both reactants is as follow,

For Benzyl chloride,
$\frac{353420 mg of Phosphonium is formed by reacting}{220 mg of Phosphonium will be formed by}$ = $\frac{126580 g of benzyl chloride}{X}$

Solving for X,
X = $\frac{220 mg . 126580 mg}{353420 mg}$
X = 78.79 mg

For PPh₃:
$\frac{353420 mg of Phosphonium is formed by reacting}{220 mg of Phosphonium will be formed by}$ = $\frac{262290 g of PPh3}{X}$

Solving for X,
X = $\frac{220 mg . 262290 mg}{353420 mg}$
X = 163.27 mg

Now, Assuming these values as for 95 % conversion, we can calculate 100 % yield as follow,

when $\frac{95 percent}{100 percent}$ = $\frac{220 g}{X}$
Solving for X,

X = $\frac{220 . 100}{95}$ = 231.57 mg

Now, calculate reactants mass with respect to 231.57 mg
when $\frac{220 mg phosphonium required}{231.57 mg require}$ = $\frac{78.79 g of benzyl chloride}{X}$
Solving for ,

X = $\frac{231.57 . 78.79}{220}$ = 82.93 mg of Benzyl chloride

when $\frac{220 mg phosphonium required}{231.57 mg require}$ = $\frac{163.27 g of PPh3}{X}$
Solving for ,

X = $\frac{231.57 . 163.27}{220}$ = 171.85 mg of PPh3

So, reaction was started with reacting 82.93 mg of Benzyl Chloride and 171.85 mg of Triphenyl Phosphine.

7 0

2 years ago

A gas occupies 72.1 at stp. At what temperature would the gas occupy 85.9 L at a pressure of 93.6 kPa?

Alika [10]

Answer:

328.1 K.

Explanation:

To calculate the no. of moles of a gas, we can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in.

If n is constant, and have two different values of (P, V and T):

P₁V₁T₂ = P₂V₂T₁

P₁ = 1.0 atm (standard P), V₁ = 72.1 L, T₁ = 25°C + 273 = 298 K (standard T).

P₂ = 93.6 kPa = 0.924 atm, V₂ = 85.9 L, T₂ = ??? K.

T₂ = P₂V₂T₁/P₁V₁ = (0.924 atm)(85.9 L)(298 K)/(1.0 atm)(72.1 L) = 328.1 K.

4 0

2 years ago

Calculate the number of grams of solute in 500.0 mL of 0.189 M KOH.

KIM [24]

Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

Solution : Given,

Volume of solution = 500 ml

Molarity of KOH solution = 0.189 M

Molar mass of KOH = 56 g/mole

Formula used :

$Molarity=\frac{\text{Mass of KOH}\times 1000}{\text{Molar mass of KOH}\times \text{Volume of solution in ml}}$

Now put all the given values in this formula, we get the mass of solute KOH.

$0.189M=\frac{\text{Mass of KOH}\times 1000}{(56g/mole)\times (500ml)}$

$\text{Mass of KOH}=5.292g$

Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

7 0

2 years ago

Read 2 more answers

How many minutes would be required to electroplate 25.0 grams of chromium by passing a constant current of 4.80 amperes through

True [87]

Answer:

483.27 minutes

Explanation:

using second faradays law of electrolysis

5 0

2 years ago