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3 years ago

4Cr(s)+3O2(g)→2Cr2O3(s) calculate how many grams of the product form when 21.4 g of O2 completely reacts

Chemistry

1 answer:

weqwewe [10]3 years ago

6 0

Answer:

= 67.79 g

Explanation:

The equation for the reaction is;

4Cr(s)+3O2(g)→2Cr2O3(s)

The mass of O2 is 21.4 g, therefore, we find the number of moles of O2;

moles O2 = 21.4 g / 32 g/mol

=0.669 moles

Using mole ratio, we get the moles of Cr2O3;

moles Cr2O3 = 0.669 x 2/3

=0.446 moles

but molar mass of Cr2O3 is 151.99 g/mol

Hence,

The mass Cr2O3 = 0.446 mol x 151.99 g/mol

 = 67.79 g 

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2 years ago

For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

Answer: The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For Iron:

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

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Read 2 more answers

In a 100 g sample of iron(III) oxide, how many grams of iron are present?

e-lub [12.9K]

The formula of Iron(III) oxide is Fe2O3
In order to calculate the mass of iron in a given sample of iron(III) oxide, we must first know the mass percentage of iron in iron(III) oxide. This is calculated by:
[mass of iron in one mole of iron(III) oxide/ mass of one mole of iron(III) oxide] * 100
= [(moles of iron * Mr of iron) / (moles of Iron * Mr of Iron + moles of Oxygen * Mr of Oxygen)] * 100
= [(2 * 56) / (2 * 56 + 3 * 16)] * 100
= (112 / 160) * 100
= 70%
Thus, in a 100g sample, the weight of iron will be:
100 * 70%
= 70 grams

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3 years ago