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2 years ago

Consider the chemical reaction

Chemistry

1 answer:

finlep [7]2 years ago

6 0

Explanation:

aC2H6 + bCO2 + cH2O → dC2H5OH

The balanced equation for the reaction is given as;

6C2H6 + 2CO2 + 3H2O → 7C2H5OH

Check!

Oxygen

Reactant = 2 * 2 + 3 = 7

Product = 7

Carbon

Reactant = 6 * 2 + 2 = 14

Product = 7 * 2 =14

Hydrogen

Reactant = 6 * 6 + 3 * 2 = 42

Product = 7 * 5 + 7 * 1 = 42

a = 6

b = 2

c = 3

d = 7

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Is a Animal and Plant cell living

Mila [183]

Yes Animal and plant are cell living

7 0

3 years ago

What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL o

Nina [5.8K]

Answer:

The concentration of NaCl = 0.3374 M

Explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

Volume of AgNO₃ = 20.22 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 20.22×10⁻³ L

Molarity of a solution is the number of moles of solute present in 1 L of the solution.

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

The formula can be written for the calculation of moles as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Thus,

$Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}$

$Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles$

$Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles$

The chemical reaction taking place:

$AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)$

According to reaction stoichiometry:

1 mole of AgNO₃ reacts with 1 mole of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 15.00×10⁻³ L

Applying in the formula of molarity as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}$

$Molarity\ of\ NaCl= 0.3374 M$

Thus, the concentration of NaCl = 0.3374 M

6 0

2 years ago

LIMITING REACTANT!! Please help I’m very confused.

alexira [117]

Answer:

We'll have 1 mol Al2O3 and 3 moles H2

Explanation:

Step 1: data given

Numer of moles of aluminium = 2 moles

Number of moles of H2O = 6 moles

Step 2: The balanced equation

2Al + 3H2O → Al2O3 + 3H2

Step 3: Calculate the limiting reactant

For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2

Aluminium is the limiting reactant. It will completely be consumed (2 moles).

H2O is in excess. There will react 3/2 * 2 = 3 moles

There will remain 6 - 3 = 3 moles

Step 4: Calculate moles products

For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2

For 2 moles Al we'll have 2/1 = 1 mol Al2O3

For 2 moles Al We'll have 3/2 * 2 = 3 moles H2

We'll have 1 mol Al2O3 and 3 moles H2

8 0

3 years ago

The vapor pressure of water at 25.0°c is 23.8 torr. determine the mass of glucose (molar mass = 180 g/mol) needed to add to 500.

svp [43]

Q: A
according to this formula, we can get the mole fraction of water (n):
P(solu) = n Pv(water)
when we have Pv(solu) = 22.8 and Pv(water) = 23.8 so by substitution:
22.8 = n * 23.8
n= 0.958
- we need to get the moles of glucose:
moles of water = 500 g(mass weight) / 18 (molar weight)= 27.7 mol
n = moles of water / ( moles of water + moles of glucose)
0.958 = 27.7 / ( 27.7+ moles of glucose)
0.958 moles of glucose + 26.5 = 27.7
0.968 moles of glucose = 1.2
moles of glucose = 1.253 mol
∴ the mass of glucose = no.of glucose moles x molar mass
= 1.253 x 180 = 225.5 g
Q: B
here we also need to get n (mole fraction of water )by using this formula:
Pv(solu) = n Pv(water)
when we have Pv(solu)=132 & Pv(water)=150 so, by substition:
132= n * 150
n = 0.88
so, mole fraction of solution = 1 - 0.88 = 0.12
and we can get after that the moles of water = (mass weight / molar mass)
- no.moles of water = 85 g / 18 g/mol = 4.7 moles
- total moles in solution = moles of water / moles fraction of water
= 4.7 / 0.88 = 5.34 moles
∴ moles of the solution = total moles in solu - moles of water
= 5.34 - 4.7 = 0.64 moles solute
∴ the molar mass of the solute = mass weight of solute / no.of moles of solute
= 53.8 / 0.64 = 84 g/mole

Q: C

moles of urea (NH2)2 CO = mass weight / molar mass
= 4.49 g / 60 g /mol
= 0.07 mol
moles of methanol = mass weight / molar mass
= 39.9 g / 32 g/mol = 1.25 mol
moles fraction of methanol = moles of methanol / (moles of methanol + moles of urea )
moles fraction of methanol = 1.25 / ( 1.25+0.07) = 0.95
by substitution in Pv formula we will be able to get the vapour pressure of the solu :
Pv(solu) = n P°v
Pv(solu) = 0.95 * 89 mm Hg
∴Pv(solu) = 84.55 mmHg

7 0

3 years ago

I’m very confused on how to solve this

Bond [772]

Recall that density is Mass/Volume. We are given the mL of liquid which is volume so all we need is mass now. We are given the mass of the granulated cylinder both with and without the liquid, so if we subtract them, we can get the mass of the liquid by itself. So, 136.08-105.56= 30.52g. This is the mass of the liquid. We now have all we need to find the density. So, let’s plug these into the density formula. 30.52g/45.4mL= 0.672 g/mL. This is our final answer since the problem requests the answer in g/mL, but be careful, because some problems in the future may ask for g/L requiring unit conversions. Also note that 30.52 was 4 sigfigs and 45.4 was 3 sigfigs, and so dividing them required an answer that was 3 sigfigs as well, hence why the answer is in the thousandths place

4 0

2 years ago