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3 years ago

For the following reaction, calculate how many moles of each product are formed when 4.05 g of water is used.

Chemistry

1 answer:

timurjin [86]3 years ago

7 0

Answer:

A. 0.225 mole of H₂

B. 0.113 mole of O₂.

Explanation:

We'll begin by calculating the number of mole in 4.05 g of water (H₂O). This can be obtained as follow:

Mass of H₂O = 4.05 g

Molar mass of H₂O = (2×1) + 16 = 2 + 16 = 18 g/mol

Mole of H₂O =?

Mole = mass /Molar mass

Mole of H₂O = 4.05 / 18

Mole of H₂O = 0.225 mole

Next, the balanced equation.

2H₂O —> 2H₂ + O₂

From the balanced equation above,

2 moles of H₂O produced 2 moles of H₂ and 1 mole of O₂.

A. Determination of the number of mole hydrogen produced.

From the balanced equation above,

2 moles of H₂O produced 2 moles of H₂.

Therefore, 0.225 mole of H₂O will also produce 0.225 mole of H₂.

B. Determination of the number of mole oxygen produced.

From the balanced equation above,

2 moles of H₂O produced 1 mole of O₂.

Therefore, 0.225 mole of H₂O will produce = (0.225 × 1)/2 = 0.113 mole of O₂.

Thus, 0.113 mole of O₂ is produced.

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One of the intermediates in the synthesis of glycine from ammonia, carbon dioxide, and methane is aminoacetonitrile, C2H4N2.

mart [117]

Answer:

You could expect 3.48 grams of C₂H₄N₂

Explanation:

You must start by stating the chemical equation for the reaction of ammonia, carbon dioxide, and methane to produce aminoaceto nitrile.

1. Word equation:

Ammonia + Carbon dioxide + Methane → Aminoacetonitrile + Water

2. Balanced chemical equation:

$8NH_3+5CO_2+3CH_4\rightarrow 4C_2H_4N_2 +10H_2O$

3. Convert the mass of each reactant into number of moles:

Formula:

Number of moles = mass in grams/molar mass

2.11g NH₃

Number of moles = 2.11g / 17.03g/mol = 0.124 mol NH₃

14.9g CO₂

Number of moles = 14.9g/44.01g/mol = 0.339 mol CO₂

1.75g CH₄

Number of moles = 1.75g/16.04g/mol = 0.109 mol CH₄

4. Theoretical mol ratio

From the balanced chemical equation, using the coefficientes:

$8molNH_3:5molCO_2:3molCH_4:4molC_2H_4N_2:10molH_2O$

5. Limiting reagent

The available amounts of the reactants are:

0.124 mol NH₃

0.339 mol CO₂

0.109 mol CH₄

Fom the theoretical mole ration, to react with 0.124 mol of NH₃ you would need:

0.124molNH₃ × (5molCO₂/8molNH₃) = 0.0775 mol CO₂

Since there are 0.339 moles available, this is in excess.

0.124molNH₃ × (3molCH₄/8molNH₃) = 0.0465mol CO₂

Since there are 0.109 moles available, this is in excess too.

Hence, the limiting reagent is NH₃.

6. Yield

Use the theoretical ratio:

0.124molNH₃ × (4molC₂H₄N₂ / 8molNH₃) = 0.0620 mol C₂H₄N₂

Convert to grams:

Mass = number of moles × molar mass

0..0620 mol × 56.068g/mol = 3.48 g of C₂H₄N₂ ← answer

4 0

4 years ago

Please help!!

xxTIMURxx [149]

Answer:

In an experiment, a student transferred 4.50 mL of a liquid into a pre-weighed beaker (the weight of which was determined to be 35.986 g ).

Explanation:

HOPE THIS HELPS YOU 

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6 0

3 years ago

Which formulas represent two polar molecules?

Tresset [83]

The correct answer is shown in option 3. Water and hydrochloric acid are polar molecules. These molecules are polar because of the presence of bonds that are partially ionic or polar covalent bonds. Other examples are hydrogen fluoride and ammonia.

7 0

4 years ago

you find that 7.36g of a compound has decomposed to give 6.93g of oxygen. the only other element in the compound is hydrogen. if

slavikrds [6]

First solve the moles of oxgen present in the compound

mol O = 6.93 g O ( 1 mol O / 16 g O )
mol O = 0.43 mol H

then solve the moles of hydrogen present
mol H = ( 7.36 - 6.93) g H ( 1 mol H / 1 g H)
mol H = 0.43 mol H
so the O and H are in the same mole content so the molecular formula would be OH, but the molar mass will not satisfy. so the answer would be
H2O2

8 0

3 years ago

Read 2 more answers

HURRYYY!!!!!!!!!!

vlabodo [156]

Im not sure but i think its 350. I really hope im right

8 0

3 years ago