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1 year ago

How much .05 M Hcl solution can be made by diluting 250 mL of 10 M HCl

1 answer:

6 0

0.05 * V1 = 250 * 10

 V1 = 2500 / 0.05 

 V1 = 2500 * 20 

 V1 = 50000 mL 

V1 = 50 Liters .

Hope this helped! :3

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Answer: 60.7 g of $PH_3$ will be formed.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}$

$\text{Moles of} H_2=\frac{60L}{22.4L}=2.68moles$

The balanced chemical reaction is

$P_4(s)+6H_2(g)\rightarrow 4PH_3(g)$

$H_2$ is the limiting reagent as it limits the formation of product and $P_4$ is the excess reagent.

According to stoichiometry :

6 moles of $H_2$ produce = 4 moles of $PH_3$

Thus 2.68 moles of $H_2$ will produce= $\frac{4}{6}\times 2.68=1.79moles$ of $PH_3$

Mass of $PH_3=moles\times {\text {Molar mass}}=1.79moles\times 33.9g/mol=60.7g$

Thus 60.7 g of $PH_3$ will be formed by reactiong 60 L of hydrogen gas with an excess of $P_4$

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1 year ago

sdas [7]

1 mole ----------- 114.22 g
22.05 moles ---- ?

mass octane = 22.05 x 114.22 / 1

mass octane = 2,518 g

Answer D

hope this helps!

6 0

1 year ago