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3 years ago

6

How many grams of calcium phosphate can be produced when 78.5 grams of calcium hydroxide reacts with excess phosphoric acid? Unb

alanced equation: H3PO4 + Ca(OH)2 → H2O + Ca3(PO4)2 So I balanced the equation as 2H3PO4 + 3Ca(OH)2 = 6H2O + Ca3(PO4)2 but now I am stuck

1 answer:

sasho [114]3 years ago

3 0

Once you balance the enquation you "switch partners" of the element (negative charge to positive charge)

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A solution contains one or more of the following ions: Ag , Ca2 , and Co2 . Lithium bromide is added to the solution and no prec

koban [17]

Answer:

$Ca^{2+} and Co^{2+}$

Explanation:

Given:

A solution contains one or more of the following ions such as Ag, $Ca_2$ and $Co_2$

Here the Lithium bromide is added to the solution and no precipitate forms

Solution:

Since with LiBr no precipitation takes place therefore Ag+ is absent

Here on adding $Li_2SO_4$ to it precipitation takes place.

Precipitate is as follows,

$Ca^{2+}(aq)+SO_4^2^-(aq)----->CASO_4(s)$

Thus,

$Ca^2^+$ is present

When $Li_3PO_4$ is added again precipitation takes place.

Therefore the reaction is as follows,

$Co^2^+(aq)+PO_4^3^-(aq)------>Co_3(PO_4)_2(s)$

Therefore,

$Ca^{2+} and Co^{2+}$ are present in the solution

3 0

3 years ago

5. The combustion of coal with oxygen forms CO2 and releases 94 kcal of energy.

Sonbull [250]

Ok buddy sorry buddy I just got

8 0

3 years ago

A reaction gives off 35.6 kJ how many calories is this?

tia_tia [17]

8.509 kilocalories I think

7 0

3 years ago

In the process of eukaryotic pre-mRNA splicing, how is the lariat intermediate formed?

Aleks [24]

Answer:

C

Explanation:

The Eukaryotic pre-mRNA receives a 5' cap and 3'poly(A) tail before Introns are removed and the mRNA is considered ready for translation.

4 0

2 years ago

For the following reaction, 4.21 grams of hydrogen gas are allowed to react with 10.6 grams of ethylene (C2H4) . hydrogen(g) + e

sergiy2304 [10]

Answer:a) 11.34 g of ethane $(C_2H_6)$ can be formed

b) $C_2H_4$ is the limiting reagent

c) 3.44 g of the excess reagent remains after the reaction is complete

Explanation:

To calculate the moles :

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

1. $\text{Moles of} H_2=\frac{4.21}{2}=2.10moles$

2. $\text{Moles of} C_2H_4=\frac{10.6}{28}=0.378moles$

$H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)$

According to stoichiometry :

1 mole of $C_2H_4$ require 1 mole of $H_2$

Thus 0.378 moles of $C_2H_4$ will require= $\frac{1}{1}\times 0.378=0.378moles$ of $H_2$

Thus $C_2H_4$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent.

moles of $H_2$ left = (2.10-0.378) = 1.72 moles

mass of $H_2$ left= $moles\times {\text {Molar mass}}=1.72moles\times 2g/mol=3.44g$

According to stoichiometry :

As 1 mole of $C_2H_4$ give = 1 mole of $C_2H_6$

Thus 0.378 moles of $C_2H_4$ give = $\frac{1}{1}\times 0.378=0.378moles$ of $C_2H_6$

Mass of $C_2H_6=moles\times {\text {Molar mass}}=0.378moles\times 30g/mol=11.34g$

Thus 11.34 g of ethane is formed.

4 0

3 years ago