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3 years ago

How many grams of solid Ca(OH)2 (74.1 g/mol) are required to make 500 ml of a 3 M solution?

Chemistry

1 answer:

mars1129 [50]3 years ago

3 0

Answer:

111.15 g are required to prepare 500 ml of a 3 M solution

Explanation:

In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.

Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of

(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.

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3 years ago

For the following reaction, 35.4 grams of zinc oxide are allowed to react with 6.96 grams of water . zinc oxide(s) + water(l) --

IRISSAK [1]

Answer:

$m_{Zn(OH)_2}=38.4g$

Explanation:

Hello!

In this case, for the undergoing chemical reaction:

$ZnO(s)+H_2O(l)\rightarrow Zn(OH)_2$

We evaluate the yielded moles of zinc hydroxide by each reactant as shown below:

$n_{Zn(OH)_2}^{by ZnO}=35.4gZnO*\frac{1molZnO}{81.38gZnO}*\frac{1molZn(OH)_2}{1molZnO} =0.435molZn(OH)_2\\\\n_{Zn(OH)_2}^{by H_2O}=6.96gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{1molZn(OH)_2}{1molH_2O} =0.386molZn(OH)_2$

In such a way, since the water yields a smaller amount of zinc hydroxide we conclude it is the limiting reactant so the maximum mass is computed below:

$m_{Zn(OH)_2}=0.386molZn(OH)_2*\frac{99.424 gZn(OH)_2}{1molZn(OH)_2} \\\\m_{Zn(OH)_2}=38.4g$

Because the water limits the yielded amount of zinc hydroxide.

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3 years ago

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Answer:

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3 years ago

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2 years ago

Read 2 more answers

9 How many grams of O₂ are needed to produce 15.5 g Fe₂O3 in the following reaction? Fe(s) + O₂(g) → Fe₂O3 (s)

Slav-nsk [51]

Answer:

Explanation:

so u can work out the amount of moles in FeO3 by doing mr of fe3o3 is 55.8*3+16*3=215.4

moles= mass/mr so you do 15.5g/215.4=0.0719 moles

then using 1 to 1 ratio so O2 moles is 0.0719

then use the equation mass=mole*mr

so 0.0719*16=1.15g

hope this make sense :)

8 0

3 years ago