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3 years ago

1) A 10g sample of H2(g) reacts with a 22g sample of O2(g) according to

Chemistry

1 answer:

solmaris [256]3 years ago

7 0

Answer:

H₂ is excess reactant and O₂ the limiting reactant

Explanation:

Based on the chemical reaction:

2H₂(g) + O₂(g) → 2H₂O

<em>2 moles of H₂ react per mole of O₂</em>

<em />

To find limiting reactant we need to convert the mass of each reactant to moles:

<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>

10g H₂ * (1mol / 2.016g) = 4.96 moles

<em>Moles O₂ -Molar mass: 32g/mol-:</em>

22g O₂ * (1mol / 32g) = 0.69 moles

For a complete reaction of 0.69 moles of O₂ are needed:

0.69mol O₂ * (2mol H₂ / 1mol O₂) = 1.38 moles of H₂

As there are 4.96 moles,

<h3>H₂ is excess reactant and O₂ the limiting reactant</h3>

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How many grams of lead(II) sulfate (303 g/mol) are needed to react with sodium chromate (162 g/mol) in order to produce 0.162 kg

Afina-wow [57]

Answer : The mass of $PbSO_4$ needed are, 1.515 grams.

Explanation :

First we have to calculate the mole of $PbCrO_4$ .

$\text{Moles of }PbCrO_4=\frac{\text{Mass of }PbCrO_4}{\text{Molar mass of }PbCrO_4}=\frac{0.162g}{323g/mole}=0.005mole$

Now we have to calculate the moles of $PbSO_4$ .

The balanced chemical reaction will be,

$PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4$ produced from 1 mole of $PbSO_4$

So, 0.005 mole of $PbCrO_4$ produced from 0.005 mole of $PbSO_4$

Now we have to calculate the mass of $PbSO_4$

$\text{Mass of }PbSO_4=\text{Moles of }PbSO_4\times \text{Molar mass of }PbSO_4$

$\text{Mass of }PbSO_4=0.005mole\times 303g/mole=1.515g$

Therefore, the mass of $PbSO_4$ needed are, 1.515 grams.

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3 years ago

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Answer:

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3 years ago

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irina [24]

Answer:

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3 years ago

Elements that belong to the same period/row!!

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Answer:

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Explanation:

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3 years ago

Determine if the bond between each pair of atoms is pure covalent, polar covalent, or ionic. drag the appropriate items to their

dsp73

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

Less than 0.4 then it is Non Polar Pure Covalent

Between 0.4 and 1.7 then it is Polar Covalent

Greater than 1.7 then it is Ionic

For Br and Br,

E.N of Bromine = 2.96

________

E.N Difference 0.00 (Non Polar/Pure Covalent)

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E.N of Oxygen = 3.44

E.N of Nitrogen = 3.04

________

E.N Difference 0.40 (Non Polar/Pure Covalent)

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E.N of Hydrogen = 2.20

E.N of Phosphorous = 2.19

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E.N of Oxygen = 3.44

E.N of Potassium = 0.82

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4 years ago