All categories

3 years ago

4. Convert the following: a. 4g mol of MgCl2 to g b. 2 lb mol of C3H8 to g c. 16 g of N2 to lb mol d. 3 lb of C2H6O to g mol

Chemistry

1 answer:

Nuetrik [128]3 years ago

6 0

Answer:

a) 381.2 g

b) 39916 g

c) 0.0013 lb mol

d) 29.6 g mol

Explanation:

The molecular weight (mw) of a compound is the mass of it per mole, so it's the ratio of the mass (m) per mole (n).

a) The molecular weight of one mol is found at the periodic table. So, for Mg, mw = 24.3 g/mol, for Cl = 35.5 g/mol, so for MgCl2, mw = 24.3 + 2*35.5 = 95.3 g/mol. The g mol is the mass divided by the molecular weight:

g mol = m/mw

4 = m/95.3

m = 381.2 g

b) The pound (lb) is a unity of mass, and the lb mol is a unity of the mass divided by the molecular weight. So, by the periodic table, the molecular weight of C3H8 is 3*12 (of C) + 8*1 (of H) = 44 lb/mol.

lb mol = m/mw

2 = m/44

m = 88 lb

1 lb = 453.592 g

So, m = 88*453.592 = 39916 g

c) The molecular weight of N2 is 2*14 (of N) = 28 lb/mol.

m = 16/453.592 = 0.0353 lb

lb mol = m/mw

lb mol = 0.0353/28

lb mol = 0.0013 lb mol

d) The molecular weight is 2*12 (of C) + 6*1(of H) + 1*16(of O) = 46 g/mol

3 lb = 1360.78 g

g mol = m/mw

g mol = 1360.78/46

g mol = 29.6 g mol

You might be interested in

The process that makes atmospheric nitrogen available to plants by mutualistic and free-living bacteria is called

artcher [175]

Nitrogen fixation is the process that makes atmospheric nitrogen available to plants by mutualistic and free-living bacteria. The process is undertaken by the rhizobium bacteria that live in root roots of plants such as legumes. The mutualistic relationship is that the plant supplies the bacteria with a habitat in which to live, water, and nutrients, and the bacteria supply nitrogen for making plant proteins.

3 0

4 years ago

(f) what is the observed rotation of 100 ml of a solution that contains 0.01 mole of d and 0.005 mole of l? (assume a 1-dm path

never [62]

<span>Answer: .01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L. so .01/.015 to .005/.015 ~ 67% D to 33% L. And thus, the enantiomer excess will be 34%.</span>

4 0

3 years ago

Read 2 more answers

The end point indicates the completion of the titration by physical change (which is

saveliy_v [14]

Your answer would be, true.

8 0

3 years ago

Decane (C10H22) is used in diesel. The combustion for decane follows the equation: 2 C10H22 + 31 O2 à 20 CO2 + 22 H2O. Calculate

creativ13 [48]

The mass of water produced is 792 grams by the combustion of 568 grams of decane.

Given:

Combustion of 568 grams of decane with 2979 grams of oxygen.

$2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O$

To find:

The mass of water produced by combustion of 568 grams of decane.

Solution:

Mass of decane = 568 g

Moles of decane :

= $\frac{568 g}{142 g/mol}=4 mol$

Mass of oxygen gas = 2976 g

Moles of oxygen gas:

= $\frac{2976 g}{32 g/mol}=93 mol$

$2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O$

According to reaction, 2 moles of decane reacts with 31 moles of oxygen, then 4 moles of decane will react with:

$=\frac{31}{2}\times 4mol=62\text{ mol of}O_2$

But according to the question, we have 93.0 moles of oxygen gas which is more than 62 moles of oxygen gas.

So, this means that oxygen gas is present in an excessive amount. Which simply means:

Oxygen gas is an excessive reagent.
Decane is a limiting reagent.
Decane being limiting reagent will be responsible for the amount of water produced after the reaction.

According to reaction, 22 moles of water is produced from 2 moles of decane, then 4 moles of decane will produce:

$=\frac{22}{2}\times 4mol=44\text{mol of }H_2O$