Is a theory a series of questions

In the reaction BaCO3 + 2HNO3 + Ba(NO3)2 + CO2 + H2O, what mass of Ba(NO3)2 can be formed by combining 55 g BaCO3 and 26 g HNO3

Nataliya [291]

From the stoichiometry of the reaction, the mass of barium nitrate produced is 54.9 g.

<h3>Stoichiometry</h3>

The term stoichiometry refers to mass - volume relationships. Stoichiometry can be used to calculate the amount, mass or volume of reactants and products from the balanced reaction equation.

The equation of the reaction is written as follows;

BaCO3 + 2HNO3 ------> Ba(NO3)2 + CO2 + H2O

Number of moles of BaCO3 = 55 g/197.34 g/mol = 0.28 moles

Number of moles of HNO3 = 26 g/63.01 g/mol = 0.41 moles

From the reaction equation;

1 mole of BaCO3 reacts with 2 moles of HNO3

0.28 moles of BaCO3 reacts with 0.28 moles × 2 moles/1 mole = 0.56 moles

There is not enough HNO3 hence it is the limiting reactant.

Number of moles of Ba(NO3)2 produced is obtained from;

2 moles of HNO3 yields 1 mole of Ba(NO3)2

0.41 moles of HNO3 yields 0.41 moles × 1 mole/2 moles

= 0.21 moles of Ba(NO3)2

Mass of Ba(NO3)2 = 0.21 moles × 261.337 g/mol = 54.9 g

Learn more about stoichiometry: brainly.com/question/9743981

8 0

3 years ago

Q#5 Balance and list the coefficients from reactants to products.

skad [1K]

A. $2Fe_2O_3(s) +3C(s)$ → $4Fe(s) + 3CO_2(g)$

B. $2Al(s) + 3FeO(s)$ → $Fe(s) + 3Al_2O_3(s)$

C. $2Al(s) +3H_2SO_4(aq)$ → $Al_2(SO_4)_3(aq) + 3H_2(g)$

What is a balanced chemical equation?

An equation that has an equal number of atoms of each element on both sides of the equation is called a balanced chemical equation.

A. $2Fe_2O_3(s) +3C(s)$ → $4Fe(s) + 3CO_2(g)$

B. $2Al(s) + 3FeO(s)$ → $Fe(s) + 3Al_2O_3(s)$

C. $2Al(s) +3H_2SO_4(aq)$ → $Al_2(SO_4)_3(aq) + 3H_2(g)$

Learn more about the balanced chemical equation here:

brainly.com/question/15052184

#SPJ1

8 0

2 years ago

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The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago

A intramuscular medication is given at 5.00mg/kg of body weight. What is the dose in grams for a 180-lb patient?

Alex17521 [72]

0.408 gram for 180-lb patient

7 0

4 years ago

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a known mass of solute is dissolved in a known mass of water and the total volume is determined. The molar mass of the solute ca

kotykmax [81]

Explanation:

Properties of a solution which are dependent on the ratio of number of solute particles to the number of solvent molecules in a solution are known as colligative properties.

Lowering of vapor pressure and elevation in boiling point are basically two of the colligative properties which indirectly help to measure the molecular weight of a substance.

Thus, we can conclude that molar mass of the solute can be determined by measuring lowering of vapor pressure and elevation in boiling point.

5 0

4 years ago