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Lina20 [59]
3 years ago
6

what is the limiting reactant when 1.50g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advan

ced batteries, according to the following unbalanced equation?
Chemistry
1 answer:
PSYCHO15rus [73]3 years ago
7 0
<h3>Answer:</h3>

Limiting reactant is Lithium

<h3>Explanation:</h3>

<u>We are given;</u>

  1. Mass of Lithium as 1.50 g
  2. Mass of nitrogen is 1.50 g

We are required to determine the rate limiting reagent.

  • First, we write the balanced equation for the reaction

6Li(s) + N₂(g) → 2Li₃N

From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.

  • Second, we determine moles of Lithium and nitrogen given.

Moles = Mass ÷ Molar mass

Moles of Lithium

Molar mass of Li = 6.941 g/mol

Moles of Li = 1.50 g ÷ 6.941 g/mol

                   = 0.216 moles

Moles of nitrogen gas

Molar mass of Nitrogen gas is 28.0 g/mol

Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol

                                     = 0.054 moles

  • According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
  • Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
  • On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.

Thus, Lithium is the limiting reagent while nitrogen is in excess.

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When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.
Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2

And the resulting percent yield:

Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%

Regards!

4 0
3 years ago
What is the molar mass of glucose, C3H603?
ziro4ka [17]

Answer:

The mass percentage of carbon can be found easily using the molar mass of C6H12O6, 180.1559 g/mol. We need to find the mass of the glucose produced, so we multiply the number of moles of glucose by its molar mass. C6H12O6 = CO2 + C3H6O3 + CH3OCH3 Take fructose for example. Compound.

Explanation: I looked it up

4 0
2 years ago
A bicyclist travels the first 800 m of a trip with a speed of 10 m/s, the next 500 m with an
bezimeni [28]

Answer:9.18 m/s

Explanation:

The average speed for the entire trip is found ......total distance/total time. Remember r*t=d, so divide both sides by t and get r = d/t.

So the cyclist went 800+500+1200 m = 2500 m for total distance.

10t =800 leads to t=80 sec

5t=500 leads to t=100 sec

13t = 1200 leads to t = 92.3 sec

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7 0
3 years ago
How much heat, in calories, is needed to raise the temperature of 125.0 g of Lead (c lead = 0.130 J/g°C) from 17.5°C to 41.1°C?
pav-90 [236]
95.6 cal
are needed.
Explanation:
Use the following equation:
q
=
m
c
Δ
T
,
where:
q
is heat energy,
m
is mass,
c
is specific heat capacity, and
Δ
T
is the change in temperature.
Δ
T
=
T
final
−
T
initial
Known
m
=
125 g
c
Pb
=
0.130
J
g
⋅
∘
C
T
initial
=
17.5
∘
C
T
final
=
42.1
∘
C
Δ
T
=
42.1
∘
C
−
17.5
∘
C
=
24.6
∘
C
Unknown
q
Solution
Plug the known values into the equation and solve.
q
=
(
125
g
)
×
(
0.130
J
g
⋅
∘
C
)
×
(
24.6
∘
C
)
=
400. J

(rounded to three significant figures)
Convert Joules to calories
1 J
=
0.2389 cal
to four significant figures.
400
.
J
×
0.2389
cal
1
J
=
95.6 cal

(rounded to three significant figures)
95.6 cal
are needed.
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What are two versions of genes called and how are they diffrent?
Stolb23 [73]

Answer with Explanation:

Different versions of a gene are called alleles. Alleles are described as either dominant or recessive depending on their associated traits. Since human cells carry two copies of each chromosome they have two versions of each gene.

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