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marysya [2.9K]
3 years ago
15

what is the balanced equation for when an aqueous solution of hydrochloric acid reacts with sodium metal, aqueous sodium chlorin

e and hydrogen gas are formed ​
Chemistry
1 answer:
FinnZ [79.3K]3 years ago
6 0

Answer:

2Na(s) + 2HCl(aq) —> 2NaCl(aq) + H2(g)

Explanation:

Na(s) + HCl(aq) —> NaCl(aq) + H2(g)

Writing an ionic equation will actually help us to understand the equation and also to balance it. This is illustrated below:

Na + H+Cl-

Na is higher than H in the activity series and as such, it will displaces H from the solution and form NaCl with H2 liberated as shown below

Na + H+Cl- —> Na+Cl- + H2

Now, put 2 in front of Na, H+Cl- and Na+Cl- to balance the equation as shown below:

2Na + 2H+Cl- —> 2Na+Cl- + H2

Now we can write the elemental equation as follow:

2Na(s) + 2HCl(aq) —> 2NaCl(aq) + H2(g)

You might be interested in
A mixture of CS2(g) and excess O2(g) is placed in a 10 L reaction vessel at 100.0 ∘C and a pressure of 3.10 atm . A spark causes
ziro4ka [17]

Answer:

PCO2  = 0.6 25 atm

PSO2  = 1.2 75 atm

PO2 = 0.6  atm

Explanation:

Step 1: Data given

Volume = 10.0 L

Temperature = 100.0 °C

Pressure = 3.10 °C

After reaction, the temperature returns to 100.0 ∘C, and the mixture of product gases (CO2, SO2, and unreacted O2) is found to have a pressure of 2.50 atm

Step 2: The balanced equation

CS2(g)+3O2(g)→CO2(g)+2SO2(g)

Step 3: Name the reactants and products

a = CS2

b = O2 before reaction

c = CO2

d = SO2

e = nS O2 after reaction with n = the number of moles

Step 4: Calculate moles before reaction

PV = nRT

n = PV/(RT)

(na + nb) = (3.10atm) * (10.0L) / ((0.08206 Latm/moleK) * (373.15K))

(na + nb) = 1.0124

Step 5: Calculate moles after reaction

PV = nRT

n = PV/(RT)

nc + nd + ne) = PV/(RT) = (2.50 atm)*(10.0L) / ((0.08206 Latm/moleK)*(373.15K))

(nc + nd + ne) = 0.816 moles

Step 6: Calculate mol fraction

For  1 mole CS2 we need 3 moles O2  to produce 1 mole of CO2 and 2 moles of SO2

moles O2 remaining = ne = nb - 3na

moles CO2 produced = nc = na

moles SO2 producted = nd = 2na

(nc + nd + ne) = 0.816 moles = nb - 3na + na + 2na = 0.816

nb = 0.816

. (na + nb) = 1.0124

na = 1.0124 moles - 0.816 moles = 0.208

which leads to  

nc = na = 0.208

nd = 2na = 2*0.208 = 0.416

ne = 0.816 - 3*0.208 = 0.192

mole fraction CO2 = 0.208 / (0.208 + 0.416 + 0.192) = 0.25

mole fraction SO2 = 0.416 / (0.208 + 0.416 + 0.192) = 0.5 1

mole fraction O2 = 0.192 /(0.208 + 0.416 + 0.192) = 0.24

Step 6: Calculate partial pressure

PCO2 = 0.25 * 2.50 atm = 0.6 25 atm

PSO2 = 0.51 * 2.50 atm = 1.2 75 atm

PO2 = 0.24 * 2.50 atm = 0.6  atm

Step 7: Control results

now let's verify a couple of things

PV = nRT

P = nRT/V

before rxn

P = (0.208 + 0.816) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 3.10 atm

after rxn

P = ((0.208 +0.416+0.192) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 2.50 atm

8 0
2 years ago
Diatomic O2 can react with the element magnesium to form magnesium oxide (MgO). The balanced chemical equation is: 2Mg + O2 → 2M
tigry1 [53]

Answer:

We will expect 4 moles of MgO to be formed (option b).

Explanation:

Step 1: The balanced equation

2Mg + O2 → 2MgO

Step 2: Data given

Number of moles of Magnesium = 4 moles

Oxygen = in excess  → this means Magnesium is the limiting reactant

Magnesium will completely be consumed ( 4 moles). There will remain 0 moles.

For 2 moles of magnesium consumed, we need 1 mole of oxygen to produce 2 moles of MgO.

For 4 moles of magnesium, we need 4/2 = 2 moles of oxygen.

For 4  moles of magnesium, we will produce 4/1 = 4 moles of MgO

We will expect 4 moles of MgO to be formed (option b).

4 0
3 years ago
Balance this equation and determine what the coefficients should be.
11Alexandr11 [23.1K]

Answer:

1, 2, 1, 1

Explanation:

3 0
3 years ago
The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentrati
a_sh-v [17]

Answer : The concentration of NOBr after 95 s is, 0.013 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.80M^{-1}s^{-1}

t = time taken  = 95 s

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.86 M

Now put all the given values in above equation, we get:

0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M

4 0
2 years ago
Which statement explains whether NaCl or BeO will have a stronger bond?
Alex777 [14]

Answer:

C

Explanation:

8 0
2 years ago
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