All categories

3 years ago

Given the problem below, what would the first step be in solving it?

Chemistry

1 answer:

Hoochie [10]3 years ago

4 0

Answer:

First step would be convert to moles

Final Answer: 37.8 g of NaCl

Explanation:

The reaction is:

2Na + Cl₂ → 2NaCI

We convert the mass of each reactant to moles:

18 g . 1mol /23g = 0.783 moles of Na

23g . 1mol / 70.9g = 0.324 moles of chlorine

We use the mole ratio to determine the limiting reactant:

Ratio is 2:1. 2 moles of Na react to 1 mol of chlorine

Then, 0.783 moles of Na, may react to (0.783 . 1)/2 = 0.391 moles.

Excellent!. We need 0.391 moles of Cl₂ and we only have 0.324 moles available. That's why the Cl₂ is our limiting reactant.

We use the mole ratio again, with the product side. (1:2)

1 mol of Cl₂ can produce 2 moles of NaCl

Then, our 0.324 moles of gas, may produce (0.324 . 2)/1 = 0.648 moles

Finally, we convert the moles to grams:

0.648 mol . 58.45g/mol =

You might be interested in

The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−

barxatty [35]

Answer:

The reaction will be non spontaneous at these concentrations.

Explanation:

$AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)$

Expression for an equilibrium constant $K_c$ :

$K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]$

Solubility product of the reaction:

$K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}$

Reaction between Gibb's free energy and equilibrium constant if given as:

$\Delta G^o=-2.303\times R\times T\times \log K_c$

$\Delta G^o=-2.303\times R\times T\times \log K_{sp}$

$\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]$

$\Delta G^o=69,117.84 J/mol=69.117 kJ/mol$

Gibb's free energy when concentration $[Ag^+] = 1.0\times 10^{-2} M$ and $[Br^-] = 1.0\times 10^{-3} M$

Reaction quotient of an equilibrium = Q

$Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}$

$\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)$

$\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])$

$\Delta G=40.588 kJ/mol$

For reaction to spontaneous reaction: $\Delta G$ .
For reaction to non spontaneous reaction: $\Delta G>0$ .

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

5 0

3 years ago

A sample of hydrated magnesium sulfate (MgSO4)

yaroslaw [1]

Answer:

MgSO4.7H2O

Explanation:

Let the formula for the hydrated magnesium sulphate be MgSO4.xH2O

Mass of the hydrated salt (MgSO4.xH2O) = 12.845g

Mass of anhydrous salt (MgSO4) = 6.273g

Mass of water molecule(xH2O) = Mass of the hydrated salt — Mass of anhydrous salt = 12.845 — 6.273 = 6.572g

Now,we can obtain the number of mole of water molecule present in the hydrated salt as follows:

Molar Mass of hydrated salt (MgSO4.xH2O) = 24 + 32 + (16x4) + x(2 + 16) = 24 + 32 + 64 + x(18) = 120 + 18x

Mass of xH2O/ Molar Mass of MgSO4.xH2O = Mass of water / mass of hydrated salt

18x/120 + 18x = 6.572/12.845

Cross multiply to express in linear form

18x x 12.845 = 6.572(120 + 18x)

231.21x = 788.64 + 118.296x

Collect like terms

231.21x — 118.296x = 788.64

112.914x = 788.64

Divide both side by 112.914

x = 788.64 /112.914

x = 7

Therefore the formula for the hydrated salt (MgSO4.xH2O) is MgSO4.7H2O

6 0

3 years ago

One of the few xenon compounds that form is cesium xenon heptafluoride (CsXeF7). How many moles of CsXeF7 can be produced from t

larisa [96]

The balanced chemical equation is written as:

CsF(s) + XeF6(s) ------> CsXeF7(s)

We are given the amount of cesium fluoride and xenon hexafluoride used for the reaction. We need to determine first the limiting reactant to proceed with the calculation. From the equation and the amounts, we can say that the limiting reactant would be cesium fluoride. We calculate as follows:

11.0 mol CsF ( 1 mol CsXeF7 / 1 mol CsF ) = 11.0 mol CsXeF7

6 0

3 years ago

Read 2 more answers

(Pls help i will give brainlyest) Label the missing parts of the diagram.

WARRIOR [948]

Answer:

Sedimentary rock

Metamorphic rock Igneous rock