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3 years ago

Given the reaction: 4A1 + 302 - 2A1,O, How many grams of Al will be

Chemistry

1 answer:

dlinn [17]3 years ago

4 0

Answer:

The answer to your question is 27 g of Al

Explanation:

Data

mass of Al = ?

moles of Al₂O₃ = 0.5

The correct formula for the product is Al₂O₃

Balanced chemical reaction

4Al + 3O₂ ⇒ 2Al₂O₃

Process

1.- Calculate the molar mass of the product

Al₂O₃ = (27 x 2) + (16 x 3)

= 54 + 48

= 102 g

2.- Convert the moles of Al₂O₃ to grams

102 g ---------------- 1 mol

x ---------------- 0.5 moles

x = (0.5 x 102) / 1

x = 51 g of Al₂O₃

3.- Use proportions to calculate the mass of Al

4(27) g of Al --------------- 2(102) g of Al₂O₃

x --------------- 51 g

x = (51 x 4(27)) / 2(102)

x = 5508 / 204

x = 27 g of Al

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3 years ago

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1) How many molecules are there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mm Hg?

RSB [31]

Answer : The number of molecules present in nitrogen gas are, $3.48\times 10^{13}$

Explanation :

First we have to calculate the moles of nitrogen gas by using ideal gas equation.

$PV=nRT$

where,

P = Pressure of $N_2$ gas = $1.00\times 10^{-6}mmHg=1.32\times 10^{-9}atm$ (1 atm = 760 mmHg)

V = Volume of $N_2$ gas = 985 mL = 0.982 L (1 L = 1000 mL)

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $0.0^oC=273+0.0=273K$

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

$(1.32\times 10^{-9}atm)\times 0.982L=n\times (0.0821L.atm/mol.K)\times 273K$

$n=5.78\times 10^{-11}mol$

Now we have to calculate the number of molecules present in nitrogen gas.

As we know that 1 mole of substance contains $6.022\times 10^{23}$ number of molecules.

As, 1 mole of $N_2$ gas contains $6.022\times 10^{23}$ number of molecules

So, $5.78\times 10^{-11}$ mole of $N_2$ gas contains $(5.78\times 10^{-11})\times (6.022\times 10^{23})=3.48\times 10^{13}$ number of molecules

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8 0

3 years ago

if you mixed 72.9g hydrochloric acid(aq) with 150g silver acetate(aq), what would be the limiting reagent?

horsena [70]

Answer:

Silver Acetate would be the Limiting Reagent.

Explanation:

The balance chemical equation for the given double displacement reaction is as;

HCl + AgC₂H₃O₂ → AgCl + HC₂H₃O₂

Step 1: Calculate Moles of Starting Materials:

Moles of HCl:

Moles = Mass / M.Mass

Moles = 72.9 g / 36.46

Moles = 1.99 moles

Moles of AgC₂H₃O₂:

Moles = 150 g / 166.91 g/mol

Moles = 0.898 moles

Step 2: Find out Limiting reagent as:

According to balance chemical equation.

1 mole of HCl reacts with = 1 mole of AgC₂H₃O₂

So,

1.99 moles of HCl will react with = X moles of AgC₂H₃O₂

Solving for X,

X = 1.99 mol × 1 mol / 1 mol

X = 1.99 mol of AgC₂H₃O₂

Hence, to completely consume 1.99 moles of Hydrochloric acid we will require 1.99 moles of Silver Acetate, But, we are provided with only 0.898 moles of Silver Acetate. This means Silver Acetate will consume first in the reaction therefore, it is the LIMITING REAGENT.

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3 years ago

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Answer:

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Explanation:

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