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3 years ago

9. How many grams of zinc are needed to produce 12 grams of zinc chloride according to the following equation? Y-

Chemistry

1 answer:

tensa zangetsu [6.8K]3 years ago

4 0

Given :

A balanced chemical equation :

$Zn +Cl_2->ZnCl_2$

To Find :

How many grams of zinc are needed to produce 12 grams of zinc chloride.

Solution :

Moles of $ZnCl_2$ ,

$n=\dfrac{Given \ wt}{Molecular\ Mass}\\\\n =\dfrac{12}{136.30}\ mol\\\\n=0.088\ mol$

Now, by balanced chemical equation we can say that 1 mol of Zn produce

1 mol of $ZnCl_2$ .

So, 0.088 mol of Zn is required to produced 0.088 mol of $ZnCl_2$ .

$Mass \ required = molecular \ mass \times moles\\\\m = 65.38 \times 0.088\\\\m=5.8 \ gm$

Therefore, 5.8 grams of zinc is required.

Hence, this is the required solution.

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The titration of a 20.0-mLmL sample of an H2SO4H2SO4 solution of unknown concentration requires 22.87 mLmL of a 0.158 M KOHM KOH

kkurt [141]

Answer:

0.0905 M

Explanation:

Let's consider the neutralization reaction between H2SO4 and KOH.

H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O

22.87 mL of 0.158 M KOH react. The reacting moles of KOH are:

0.02287 L × 0.158 mol/L = 3.61 × 10⁻³ mol

The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 3.61 × 10⁻³ mol = 1.81 × 10⁻³ mol

1.81 × 10⁻³ moles of H₂SO₄ are in 20.0 mL. The molarity of H₂SO₄ is:

M = 1.81 × 10⁻³ mol / 0.0200 L = 0.0905 M

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3 years ago

What is the mass of 0.513 mol Al2O3? Give your answer to the correct number of significant figures. (Molar mass of Al2O3 = 102.0

prisoha [69]

.513mol x (102g/1mol)

Essentially, this is .513 x 102
Which equals: 52.326
But because you can only have 3 significant figures, your answer is:
52.3 grams

I hope this Helps!

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3 years ago

Read 2 more answers

The gram formula mass of NH4Cl is(1) 22.4 g/mole (3) 53.5 g/mole(2) 28.0 g/mole (4) 95.5 g/mole

g100num [7]

The answer is (3) 53.5 g/mol. The gram formula mass means that the mass of one mol compound. 1N=14, 4H=4, 1Cl=35.5. So the gram formula mass of NH4Cl=14+4+35.5=53.5 g/mol.

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3 years ago

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A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

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

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

Help ASAP 15 points

garri49 [273]

Answer: number 1

Explanation:

6 0

2 years ago