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vova2212 [387]
3 years ago
11

PLEASE HELP!!! 890 g of reactant A completely reacts with 120 g of reactant B to form two products. If product A has a mass of 9

2 g, calculate the mass of product B based on the law of conservation of mass
Chemistry
1 answer:
ohaa [14]3 years ago
7 0

it's (890+120)-92 which is 918g

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In a laboratory experiment the reaction of 3.0 mol of H2 with 2.0 mol of I2 produced 1.0 mole of HI. Determine theoratical yield
Kobotan [32]

Answer:

Theoretical yield of HI is 512 g.

The percent yield for this reaction is 25%.

Explanation:

H_2+I_2\rightarrow 2HI

Moles of hydrogen gas = 3.0 moles

Moles of iodine gas  = 2.0 moles

According to reaction 1 mol of hydrogen gas reacts with 1 mol of iodine gas.

Then 3.0 moles of hydrogen gas reacts with 3.0 mol of iodine gas. But there are 2.0 moles of iodine gas. Hence,Iodine is a limiting reagent. The production of HI will depend upon iodine gas moles.

According to reaction , 1 mol of iodine gas gives 2 moles of HI.

Then 2 moles of iodine gas will give:

\frac{2}{1}\times 2 mol=2 mol of HI

Theoretically we will get 4 moles of HI.

Theoretical yield of HI =  4 mol × 128 g/mol= 512 g

Experimental yield of HI = 1.0 mol

= 1 mol × 128 g/mol= 128 g

\%yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}} \times 100

\%yield=\frac{128 g}{512 g}\times 100=25\%

The percent yield for this reaction is 25%.

6 0
3 years ago
What volume of a 3.5 m Hcl is required to completely neutralize 50.0ml of a 2.0 m naoh
Sauron [17]

Hey there!:

Molarity HCl = 3.5 M

Volume HCl = ?

Molarity NaOH = 2.0 M

Volume NaOH in liters = 50.0 mL / 1000 => 0.05 L

Number of moles NaOH:

n = M * V

n = 2.0 * 0.05

n = 0.1 moles of NaOH

Given the reaction:

HCl + NaOH = NaCl + H2O

1 mole HCl --------- 1 mole NaOH

1 mol HCl reacts with 1 mol NaOH , so moles NaOH = moles HCl

0.1 moles of NaOH = 0.1 moles of HCl

Therefore:

M( HCl ) = n / V

3.5 = 0.1 / V

V = 0.1 / 3.5

V = 0.029 L in mL :  0.029 * 1000 => 29.0 mL


hope that helps!


3 0
2 years ago
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