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saveliy_v [14]
3 years ago
12

Ydrogen and fluorine combine according to the equation h2(g) + f2(g) → 2 hf(g) if 5.00 g of hydrogen gas are combined with 38.0

g of fluorine gas, the maximum mass of hydrogen fluoride that could be produced is
Chemistry
1 answer:
Alexxx [7]3 years ago
4 0
<span> Molar mass (H2)=2*1.0=2.0 g/mol
Molar mass (F2)=2*19.0=38.0 g/mol
Molar mass (HF)=1.0+19.0=20.0 g/mol

5.00 g H2 * 1mol H2 /2 g H2=2.50 mol H2 
38.0 g F2*1mol F2/38.0 g F2=1.00 mol F2

                                   H2(g) + F2(g) → 2 HF(g)
From reaction        1 mol      1 mol
From problem      2.50 mol   1 .00mol

We can see that  excess of H2, and that F2 is a limiting reactant.
So, the amount of HF is limited by the amount of F2.

</span>                                 H2(g) + F2(g) → 2 HF(g)
From reaction                      1 mol       2  mol
From problem                      1.00 mol  2.00mol

2.00 mol HF can be formed.

2.00 mol HF*20.0g HF/1mol HF=40.0 g HF can be formed
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Convert oxygen+ hydrogen into formula equation and balance them​
nika2105 [10]

Explanation:

2H2(g) + O2(g) → 2H2O(l ) Chemical equations give the following information about chemical reactions. Chemical equations show the formulas for the substances that take part in the reaction.

5 0
2 years ago
An 240. mL sample of oxygen at 90.00C and was cooled at constant pressure
adoni [48]

The final temperature : T₂ = 680.625 K

<h3>Further explanation</h3>

Given

V₁=240 ml

T₁ = 90 + 273 = 363 K

V₂ = 450 ml

Required

The final temperature

Solution

Charles's Law  

When the gas pressure is kept constant, the gas volume is proportional to the temperature  

\tt \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

Input the value :

T₂ = V₂T₁/V₁

T₂ = 450 x 363 / 240

T₂ = 680.625 K

7 0
3 years ago
The half-life of Zn-71 is 2.4 minutes. If one had 100.0 g at the beginning, how many grams would be left after 7.2 minutes has e
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8 0
4 years ago
For the following reaction, 76.0 grams of barium chloride are allowed to react with 67.0 grams of potassium sulfate. barium chlo
Nataliya [291]

For the following reaction, 76.0 grams of barium chloride are allowed to react with 67.0 grams of potassium sulfate.

The reaction consumes _____ moles of barium chloride. The reaction produces _____ moles of barium sulfate and _____ moles of potassium chloride.

Answer: a) The reaction consumes 0.365 moles of barium chloride.

b) The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of barium chloride}=\frac{76.0}{208g/mol}=0.365moles

\text{Moles of potassium sulphate}=\frac{67.0}{174g/mol}=0.385moles

BaCl_2(aq)+K_2SO_4(aq)\rightarrow BaSO_4(s)+2KCl(aq)

According to stoichiometry :

1 mole of BaCl_2 require 1 mole of K_2SO_4

Thus 0.365 moles of BaCl_2 will require=\frac{1}{1}\times 0.365=0.365moles  of K_2SO_4

Thus BaCl_2 is the limiting reagent as it limits the formation of product and K_2SO_4 is the excess reagent.

As 1 moles of BaCl_2 give = 1 moles of BaSO_4

Thus 0.365 moles of BaCl_2 give =\frac{1}{1}\times 0.365=0.365moles  of BaSO_4

As 1 moles of BaCl_2 give = 2 moles of KCl

Thus 0.365 moles of BaCl_2 give =\frac{2}{1}\times 0.365=0.730moles  of KCl

Thus the reaction consumes 0.365 moles of barium chloride. The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.

6 0
4 years ago
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