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Zanzabum
2 years ago
11

Water is produced from the reaction of hydrogen and oxygen gas, according to the equation below. What is the excess reactant in

the reaction of 4.2 moles of hydrogen with 3.0 moles of oxygen?
2H2(g) + O2(g) = 2H2O(l)
Select one:
a.Water
b.Oxygen
c.No excess reactant
d.Hydrogen
Chemistry
2 answers:
insens350 [35]2 years ago
7 0

Answer:

2H2O means two molecules of water(H2O)

VashaNatasha [74]2 years ago
5 0

Answer:

b.Oxygen

this is the excess reactant

Why?

2 moles of hydrogen react with 1 mole of oxygen. So, 4.2 moles of hydrogen would need 2.1 moles of oxygen. There are 3.0 moles of oxygen. So, there are 0.9 moles of oxygen in excess.

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If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×
Elodia [21]

Answer:

2.30 × 10⁻⁶ M

Explanation:

Step 1: Given data

Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M

Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³

Step 2: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂

We will use the following expression.

Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²

[OH⁻] = 2.30 × 10⁻⁶ M

3 0
3 years ago
Consider the synthesis of ammonia 3 H2+ N2 à 2 NH3 If a student were to react 38.5 g of nitrogen gas, how many moles of ammonia
Wittaler [7]

Answer:

2.75 mol

Explanation:

Given data:

Mass of Nitrogen = 38.5 g

Moles of ammonia produced = ?

Solution:

Chemical  equation:

N₂ + 3H₂    →     2NH₃

Number of moles of nitrogen:

Number of moles = mass/ molar mass

Number of moles = 38.5 g/ 28 g/mol

Number of moles = 1.375 mol

Now we will compare the moles of ammonia and nitrogen from balance chemical equation.

           N₂            :            NH₃

            1              :             2

           1.375       :           2×1.375 = 2.75 mol

Thus 2.75 moles of ammonia  are produced from 38.5 g of nitrogen.

4 0
3 years ago
Complete combustion of 2.60 g of a hydrocarbon produced 8.46 g of co2 and 2.60 g of h2o. what is the empirical formula for the h
zubka84 [21]
1 mole of carbon dioxide contains a mass of 44 g, out of which 12 g are carbon. 
Hence, in this case the mass of carbon in 8.46 g of CO2:
 (12/44) × 8.46 = 2.3073 g
 1 mole of water contains 18 g, out of which 2 g is hydrogen;
Therefore, 2.6 g of water contains;
 (2/18) × 2.6 = 0.2889 g of hydrogen.
Therefore, with the amount of carbon and hydrogen from the hydrocarbon we can calculate the empirical formula.
We first calculate the number of moles of each,
Carbon = 2.3073/12  = 0.1923 moles
Hydrogen = 0.2889/1 = 0.2889 moles
Then, we calculate the ratio of Carbon to hydrogen by dividing with the smallest number value;
             Carbon : Hydrogen
  0.1923/0.1923 : 0.2889/0.1923
                       1 :  1.5
                      (1 : 1.5) 2
                     = 2 : 3
Hence, the empirical formula of the hydrocarbon is C2H3
5 0
3 years ago
32 gm of O2 to mole of O2
lana [24]

Molar mass of O2:-

\\ \rm\longmapsto 2(16u)=32g/mol

Now

\boxed{\sf No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}}

\\ \rm\longmapsto No\:of\;moles=\dfrac{32}{32}

\\ \rm\longmapsto No\:of\;moles=1mol

5 0
3 years ago
Will Maynez burns a 0.6-g peanut beneath 50 g of water, which increases in temperature from 22°C to 50°C. (The specific heat cap
Neko [114]

Answer:

40% of  the energy release by the peanut is 3500 calories

Explanation:

One calorie is defined as the amount of energy required to increase the temperature of one gram of water for one degree Celsius (or one Kelvin)

Equation for energy gain by water is

Q = mcΔT

where, m is the mass of the object

c is the specific heat capacity

ΔT is the change in temperature

c =  1.0 cal/g?°C.

m = 50 g

ΔT = 50°C - 22°C

    = 28°C

Q = (50)× (1)× (28)

  = 1400calories

The peanut contain 1400calories of energy .

amount that 40% of energy is released to water ,

so,

Q = 1400 calories / 0.4

= 3500 calories

Therefore, 40% of  the energy release by the peanut is 3500 calories

7 0
2 years ago
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