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3 years ago

H3PO4 +CaCOH)2=Ca3(PO4)2+H20 The equation balance

Chemistry

1 answer:

AnnyKZ [126]3 years ago

7 0

Answer:

2H3PO4 +3Ca(OH) 2 = Ca3 (PO4) 2 +6H20

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When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$

Finally, we compute the percent yield with the obtained 2.10 g:

$Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%$

Best regards.

7 0

2 years ago

Write the word and balanced chemical equations for the reaction between:

kupik [55]

Answer:

Alumminum hydroxide reacts with sulfuric acid as follows: 2Al(OH)3+H2SO4-->Al2(SO4)+6H2O.

Explanation:

plz mark as brainlist

7 0

2 years ago

Describe how a new drilling site off the coast of Alaska would be analyzed.

sveta [45]

Answer:

Explanation:

ポイントをありがとう

6 0

3 years ago

3. The dry air in the Alps helped to prevent damage to the cells in the Iceman's body.

Sever21 [200]

Answer:

False

Explanation:

because they do not prevent damage they make damage

3 0

2 years ago

How many moles of oxygen atoms are present in 30.5 grams of hydrogen peroxide (h2o2)?

Pepsi [2]

1.8 moles of oxygen atoms are present in 30.5 grams of hydrogen peroxide.

<u>Explanation:</u>

First we have to convert the given weight of hydrogen peroxide to molar mass of hydrogen peroxide. So for this, we have to divide the given weight with the molecular mass of hydrogen peroxide.

$\text {Molecular mass of } \mathrm{H}_{2} \mathrm{O}_{2}=(2 \times 1)+(2 \times 16)=2+32=34 \mathrm{g}$

So,

$\text {Molar mass of } \mathrm{H}_{2} \mathrm{O}_{2}=\frac{30.5}{34}=0.90 \text { moles of } \mathrm{H}_{2} \mathrm{O}_{2}$

Second step, in this moles, 2 molecules of oxygen are present. Thus 1 mole of Hydrogen peroxide consists of 2 moles of oxygen. Then,

$0.90 moles of $\mathrm{H}_{2} \mathrm{O}_{2}=2 \times 0.90=1.8$ moles of oxygen$

So, 30.5 grams of hydrogen peroxide consists of 1.8 moles of oxygen.

4 0

2 years ago