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dalvyx [7]
3 years ago
9

There are 4.67 * 10 ^ 24 hydrogen gas molecules the react with 21.0 Liters of oxygen gas STP conditions to form water. How many

grams of water will form? How many moles of the excess reactant will remain after the reaction is completed?
Chemistry
1 answer:
Zina [86]3 years ago
7 0

Answer:

I am almost done

Explanation:

21.0 L

↓

2H₂(g) + O₂(g) → 2H₂O(l)

↑

4.67 × 10²⁴ moles

[This is done so the equation is balanced and the reaction can occur]

[subscripts are the number of atoms, and the coefficients are the number of molecules]

______________

Since there are about 6.02 × 10²³ moles in one mol of a substance.

4.67 × 10²⁴ / 6.02 × 10²³ =

7.7547174437 mol.

Next, we must determine the grams per mole of each substance, this can be done with the periodic table.

Since hydrogen, and oxygen gas are diatomic, they will consist of two atoms each. So the atomic mass will be multiplied by 2 to get the molecular mass. Therefore H₂ will have a molecular mass of 1.00794 × 2 = 2.01588 g/mol.

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How many grams of PbBr2 will precipitate when excess CuBr2 solution is added to 77.0 mL of 0.595 M Pb(NO3)2 solution?Pb(NO3)2(aq
Alexxx [7]

Answer:

16.89g of PbBr2

Explanation:

First, let us calculate the number of mole of Pb(NO3)2. This is illustrated below:

Molarity of Pb(NO3)2 = 0.595M

Volume = 77mL = 77/1000 = 0.077L

Mole =?

Molarity = mole/Volume

Mole = Molarity x Volume

Mole of Pb(NO3)2 = 0.595x0.077

Mole of Pb(NO3)2 = 0.046mol

Convert 0.046mol of Pb(NO3)2 to grams as shown below:

Molar Mass of Pb(NO3)2 =

207 + 2[ 14 + (16x3)]

= 207 + 2[14 + 48]

= 207 + 2[62] = 207 +124 = 331g/mol

Mass of Pb(NO3)2 = number of mole x molar Mass = 0.046 x 331 = 15.23g

Molar Mass of PbBr2 = 207 + (2x80) = 207 + 160 = 367g/mol

Equation for the reaction is given below:

Pb(NO3)2 + CuBr2 —> PbBr2 + Cu(NO3)2

From the equation above,

331g of Pb(NO3)2 precipitated 367g of PbBr2

Therefore, 15.23g of Pb(NO3)2 will precipitate = (15.23x367)/331 = 16.89g of PbBr2

8 0
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Explanation:

4 0
3 years ago
What is the new concentration of 25.0mL added to 125.0mL of 0.150M
____ [38]

Answer:

The new concentration is 0.125 M.

Explanation:

Given data:

Initial volume V₁ = 125.0 mL

Initial molarity M₁ = 0.150 M

New volume V₂ = 25 mL +125 mL = 150 mL

New concentration M₂ = ?

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M₁V₁    =    M₂V₂

0.150 M × 125 mL = M₂ × 150 mL

M₂ = 0.150 M × 125 mL / 150mL

M₂ = 18.75 M.mL/150 mL

M₂ = 0.125 M

The new concentration is 0.125 M.

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Answer:

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