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AVprozaik [17]
2 years ago
14

How many moles of oxygen are required to make 4 moles of water?​

Chemistry
2 answers:
kakasveta [241]2 years ago
8 0

Answer:

2 moles of oxygen

Explanation:

Basic Equation (given) :

2H₂ (g) + O₂ ⇒ 2H₂O (g)

  • From this, we understand for every 2 moles of water, 1 mole of oxygen is needed

Taking the number of moles of oxygen and water in proportion :

  • 1 : 2
  • 1 × 2 : 2 × 2
  • 2 : 4

Therefore, 2 moles of oxygen are required to make 4 moles of water.

frozen [14]2 years ago
6 0

\textsf{\qquad\qquad\huge\underline{{\sf Answer}}}

As per the given equation, we can infer that :

For getting 2 moles mole of water, 1 mole of oxygen gas was required. so to get 4 moles of water we need :

We just have to multiply the whole reaction by 2 on both sides, in order to get 4 moles of H2O on product side.

\qquad \qquad \rm4 \: H_2 + 2 \: O_2  \rightarrow4 \: H_2O

therefore, we need

\qquad \sf  \dashrightarrow \: 2 \times 1 = 2 \: moles \:  \: oxygen \: gas

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A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = \frac{0.1028 mol}{0.1 L}=1.028 mol/L

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = M_1=1.208 mol

Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

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4 years ago
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