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Naddika [18.5K]
3 years ago
8

Hat coefficients are needed to balance the equation for the complete combustion of methane? enter the coefficients in the order

ch4, o2, co2, and h2o, respectively.
Chemistry
1 answer:
klio [65]3 years ago
7 0
The chemical equation for the combustion of methane is CH4 + 2O2 --> CO2 + 2H2O.

The coefficients of that equation are 1, 2, 1, and 2, in the order you asked for.
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0.446 g of hydrogen gas fills a 5.0 L bag determine the density of hydrogen
Oduvanchick [21]

The density of hydrogen : ρ = 0.0892 g/L

<h3>Further explanation</h3>

Given

mass of Hydrogen : 0.446 g

Volume = 5 L

Required

The density

Solution

Density is a quantity derived from the mass and volume  

Density is the ratio of mass per unit volume  

The unit of density can be expressed in g/cm³, kg/m³, or g/L

Density formula:  

\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}

Input the value :

ρ = m : V

ρ = 0.446 g : 5 L

ρ = 0.0892 g/L

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A synthesis reaction takes place when carbon monoxide (CO) and hydrogen gas (H2) react to form methanol (CH3OH). How many grams
jek_recluse [69]

The mass of methanol produced is 8.0 g.

We have the masses of two reactants, so this is a <em>limiting reactant</em> problem.

We know that we will need a <em>balanced equation</em> with masses, moles, and molar masses of the compounds involved.

<em>Step 1</em>. <em>Gather all the informatio</em>n in one place with molar masses above the formulas and everything else below them.

MM: ___28.01  2.016 ___32.04

_______CO + 2H_2 → CH_3OH

Mass/g: 7.0 __2.5

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>

Moles of CO = 7.0 g CO × (1 mol CO/28.01g CO) = 0.250 mol CO

Moles of H_2 =2.5 g H_2 × (1 mol H_2/2.016 g H_2) = 1.24 mol H_2

<em>Step 3. </em>Identify the<em> limiting reactan</em>t

Calculate the <em>moles of CH_3OH</em> we can obtain from each reactant.

<em>From CO</em>: Moles of CH_3OH = 0.250 mol CO  × (1 mol CH_3OH /1 mol CO)

= 0.250 mol CH_3OH

<em>From H_2</em>: Moles of CH_3OH = 1.24 mol H_2 × (1 mol CH_3OH /2 mol H_2)

= 0.620 mol CH_3OH

<em>CO is the limiting reactant</em> because it gives the smaller amount of CH_3OH.

<em>Step 4</em>. Calculate the <em>mass of CH_3OH</em>

Mass of CH_3OH = 0.250 mol CH_3OH × (32.04 g CH_3OH /1 mol CH_3OH) = 8.0 g CH_3OH

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