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notsponge [240]
4 years ago
14

Consider this reaction. What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP? 1860 mL 2880 m

L 3710 mL 45,100 mL
Chemistry
2 answers:
charle [14.2K]4 years ago
4 0
<span>Step 1: find number of moles of Mg(divide 4.03 by its molor mass)
step 2:find number of moles in O2
step 3: remember STP is 22.4 L = 1 mole</span>
step 4: multiply 0.0829mol O₂ by 22.4L/mol to get 1.857L
answer is 1860 mL
Sholpan [36]4 years ago
4 0

Answer: 1860 mL

Explanation:

2Mg+O_2\rightarrow 2MgO

According to Avogadro's law , 1 mole of every gas occupies 22.4 L at standard temperature and pressure and 1 mole of every substance weighs equal to its molar mass.

2moles=2\times 24.305=48.61g of Magnesium reacts with 22.4 L of oxygen gas STP.

4.03g of  Magnesium reacts with=\frac{22.4L}{48.61g}\times 4.03g=1.86L=1860ml of oxygen gas STP.

You might be interested in
How many moles are in 16 grams of nitrogen molecule
vekshin1
Answer:
≈ 1.142 moles

Explanation:
formula is “grams given/molar mass = moles”
nitrogen’s molar mass is 14.0067g
solve:
16 / 14.0067 = 1.14231046571
or ≈ 1.142 moles

hope this helped :)
7 0
2 years ago
PLEASE HELP MEEE I WILL MARK BRAINLIST FOR THIS ITS SUPER SIMPLE....BUT U MUST TYPE EVERYTHING
frozen [14]
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6 0
3 years ago
What is the formula weight (amu) of the molecule H2O? Use atomic masses of H and O as 1.008 amu and 16.00 amu respectively. Repo
vaieri [72.5K]

Answer:

Formula weight of H₂O molecule is  18.02 amu.

Explanation:

Given data:

Formula weight of H₂O = ?

Atomic mass of H = 1.008 amu

Atomic mass of O = 16.00 amu

Solution:

Formula weight:

"It is the sum of all the atomic weight of atoms present in given formula"

Formula weight of H₂O = 2×1.008 amu + 1×16.00 amu

Formula weight of H₂O = 18.02 amu

Thus, formula weight of H₂O molecule is  18.02 amu.

6 0
3 years ago
Does anyone have any idea what this means lol
miv72 [106K]

Answer:

29.42 Litres

Explanation:

The general/ideal gas equation is used to solve this question as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K

According to the information provided in this question;

mass of nitrogen gas (N2) = 25g

Pressure = 0.785 atm

Temperature = 315K

Volume = ?

To calculate the number of moles (n) of N2, we use:

mole = mass/molar mass

Molar mass of N2 = 14(2) = 28g/mol

mole = 25/28

mole = 0.893mol

Using PV = nRT

V = nRT/P

V = (0.893 × 0.0821 × 315) ÷ 0.785

V = 23.09 ÷ 0.785

V = 29.42 Litres

6 0
3 years ago
Assume that 12.0 g of oxygen are reacted with 20.0 g of magnesium to produce magnesium oxide.
FinnZ [79.3K]
Write out the eqn of magnesium and oxygen. this should be under “metals” chapter. do revise.

next, find the mols of both oxygen and magnesium. compare the ratios and find the LIMITING REAGENT.

use the mols of the limiting reagent to compare with the mols of the product.

take the mols of the product/mr of the product.

this will give u the mass.
8 0
3 years ago
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