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3 years ago

P4 +6Br2 --> 4PBr3

Chemistry

2 answers:

Sever21 [200]3 years ago

8 0

Answer:

$\boxed {\boxed {\sf 1 \ mole \ P_4 consumed \ to \ 4 \ moles \ PBr_3 \ produced}}$

Explanation:

To find the mole ratio, we must use the coefficients.

$P_4+6Br_2 \rightarrow 4PBr_3$

This equation is balanced (4 atoms of P and 12 atoms of Br on both sides), so we can go straight to the coefficients.

P₄ doesn't have a coefficient, so 1 is implied
Br₂ has a coefficient of 6
PBr₃ has a coefficient of 4

So according to the coefficients above, for every 1 mole of P₄ that is consumed, 4 moles of PBr₃ are produced. Therefore, the ratio is 1 mole P₄ to 4 moles of PBr₃.

grin007 [14]3 years ago

6 0

Answer:

1 mole of P₄ consumed:4 moles of PBr₃ produced

Explanation:

Look at the coefficients in the balanced chemical equation to determine the ratios. For every 1 mole of P₄ consumed (P₄ has a coefficient of 1), 4 moles of PBr₃ would be produced (PBr₃ has a coefficient of 4).

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Answer: The amount of carbon dioxide formed in the reaction is 5.663 grams

Explanation:

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

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

Given mass of oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{8g}{32g/mol}=0.25mol$

For the given chemical equation:

$7O_2(g)+2C_2H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)$

By Stoichiometry of the reaction:

7 moles of oxygen gas produces 4 moles of carbon dioxide

So, 0.25 moles of oxygen gas will produce = $\frac{4}{7}\times 0.25=0.143mol$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.143 moles

Putting values in equation 1, we get:

$0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.143mol\times 44g/mol)=6.292g$

To calculate the experimental yield of carbon dioxide, we use the equation:

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

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

$90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{90\times 6.292}{100}=5.663g$

Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams

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