All categories

3 years ago

How many grams of Ca(NO3)2 are needed to make 25.0 g of a 15.0% Ca(NO3)2(aq)?

Chemistry

1 answer:

yarga [219]3 years ago

5 0

Answer:

3.75 g.

Explanation:

mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.

mass % = (mass of solute/mass of solution) x 100.

mass of calcium nitrite = ??? g,

mass of the solution = 25.0 g.

∴ mass % = (mass of solute/mass of solution) x 100

∴ mass of solute (calcium nitrite) = (mass %)(mass of solution)/100 = (15.0 %)(25.0 g)/100 = 3.75 g.

You might be interested in

What is the correct ionic formula when mg2+ and p3− react?

Mumz [18]

Answer: Ionic formula will be $Mg_3P_2$ .

Explanation: $Mg^{2+}$ and $P^{3-}$ ions will form a ionic compound. Ionic compounds have both metals and non-metals.

Here $Mg^{2+}$ is a metal and $P^{3-}$ is a non-metal.

The net charge on any compound must be 0.

So we need 2 phosphate ions to balance the charge on $Mg^{2+}$ ions. Similarly we need 3 Magnesium ions to balance the charge on $P^{3-}$ ions.

Criss-crossing the charges, we will get the formula as $Mg_3P_2$

Criss-crossing is shown in the image below.

6 0

3 years ago

Read 2 more answers

20 poll

aleksley [76]

Answer: n=15.56moles

Explanation:

PV = nRT

where

P is pressure in atmospheres

V is volume in Liters

n is the number of moles of the gas

R is the ideal gas constant = given as (0.0821L -atm/k-mol

PV = nRT

n= PV/RT

n= (1.5 X 230)/ (0.0821 X 270)

n= 15.56 moles

5 0

3 years ago

What do you need in order to generate electricity using light from the sun?

Klio2033 [76]

Solar panels ( Solar-powered photovoltaic )

6 0

3 years ago

The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3)

cestrela7 [59]

Answer:

a. The limiting reactant is $NaHCO_{3}$

b. 0.73 g of carbon dioxide are formed.

c. The grams of excess reactant that do not participate in the reaction are 0333 g.

Explanation:

You know the following reaction:

$3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}$ ⇒ $3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}$

First, you determine the molar mass of each compound. For that you must take into account the atomic mass of each element:

Na: 23
H: 1
C: 12
O: 16

To determine the molar mass of each compound, you multiply the most atomic of each element present in the molecule by the sub-index that appears after each number, which indicates the present amount of each element in the compound:

$NaHCO_{3}$ :23+1+12+16*3=84 g/mol
$H_{3} C_{6} HO_{7}$ :1*3+12*6+1*5+16*7= 192 g/mol
$CO_{2}$ :12+16*2= 44 g/mol
$H_{2} O$ :1*2+16= 18 g/mol
$Na_{3} C_{6} H_{5} O_{7}$ : 23*3+12*6+1*5+16*7= 258 g/mol

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of $NaHCO_{3}$ react with 1 mole of $H_{3} C_{6} HO_{7}$ Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 192 g

You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:

If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

$grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}$

grams of sodium bicarbonate= 1.8375 g

But to perform the experiment you have only 1.4 g of sodium bicarbonate. So the limiting reagent is sodium bicarbonate.

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of $NaHCO_{3}$ react with 3 mole of $CO_{2}$ . Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 132 g

You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?

$grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}$

grams of carbon dioxide= 0.73 g

Then, 0.73 g of carbon dioxide are formed.

To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?

$grams of citric acid=\frac{1.4 g * 192 g}{252 g}$

grams of citric acid= 1.067 g

But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:

grams of excess reactant=1.4 g - 1.067 g

grams of excess reactant=0.333 g

So the grams of excess reactant that do not participate in the reaction are 0333 g.

3 0

3 years ago

Which one is ittt? help please:)

Alexus [3.1K]

Answer:

Explanation:

4 0

3 years ago