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deff fn [24]
3 years ago
6

How many grams of hydrogen are needed to produce 1.80 g of water according to this equation? 2H2 + O2 → 2H2O

Chemistry
2 answers:
denpristay [2]3 years ago
8 0

Answer : The correct option is, (C) 0.200 g

Explanation : Given,

Mass of water = 1.80 g

Molar mass of water = 18 g/mole

Molar mass of H_2 = 2 g/mole

First we have to calculate the moles of water.

\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{1.80g}{18g/mole}=0.1mole

Now we have to calculate the moles of hydrogen gas.

The given balanced reaction is,

2H_2+O_2\rightarrow 2H_2O

From the balanced reaction we conclude that

As, 2 mole of water obtained from 2 mole of H_2 gas

So, 0.1 mole of water obtained from 0.1 mole of H_2 gas

Now we have to calculate the mass of hydrogen gas.

\text{Mass of }H_2=\text{Moles of }H_2\times \text{Molar mass of }H_2

\text{Mass of }H_2=(0.1mole)\times (2g/mole)=0.2g

Therefore, the mass of hydrogen gas required will be, 0.2 grams

s344n2d4d5 [400]3 years ago
4 0
Answer is c

Work out the moles of 2H2O: 1.80/ (16+1+1) = 0.1
The moles of 2H2 are the same (0.1)
The Mr of H2 is 2

So the mass needed to produce 1.80g of water is
0.1 x 2 = 0.2
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How do we balance Zn + HNO3 Zn(NO3)2 + NO + H20​
USPshnik [31]

3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O

<h3>Further explanation </h3>

Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:  

  • 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.  
  • 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product  
  • 3. Select the coefficient of the substance with the most complex chemical formula equal to 1  

For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:  

Balancing C atoms, H and the last O atoms

Reaction

Zn + HNO₃⇒ Zn(NO₃)₂ + NO + H₂O

  • 1. gives a coefficient

aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O

  • 2. make an equation

Zn : left = a, right =1 ⇒a=1

H : left = b, right = 2d⇒ b=2d (eq 1)

N : left = b, right = 2+c⇒b=2+c (eq 2)

O : left = 3b, right = 6+c+d ⇒3b=6+c+d(eq 3)

  • From eq 1 and eq 3

3(2d)=6+c+d

6d=6+c+d

5d=6+c (eq 4)

  • From eq 2 and eq 3

3(2+c)=6+c+d

6+3c=6+c+d

2c=d (eq 5)

  • From eq 4 and eq 5

5(2c)=6+c

10c=6+c

9c=6

c = 2/3

  • input eq 5

d = 2 x 2/3

d = 4/3

  • input eq 1

b = 2 x 4/3

b = 8/3

The equation

aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O to

Zn + 8/3HNO₃⇒ Zn(NO₃)₂ + 2/3NO + 4/3H₂O x 3

3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O

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yanalaym [24]
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<span> First ionic bonds in this salt are separeted because of heat: 
</span>NaI(l) → Na⁺(l) + I⁻(l).

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2Na⁺(l) + 2e⁻ → 2Na(l).

Reaction of oxidation at anode(+): 2I⁻(l) → I₂(l) + 2e⁻.

The anode is positive and the cathode is negative.


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