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3 years ago

How many grams of hydrogen are needed to produce 1.80 g of water according to this equation? 2H2 + O2 → 2H2O

Chemistry

1 answer:

Arisa [49]3 years ago

7 0

Hey there!:

Molar mass:

H2 = 2.01 g/mol ; H2O = 18.01

Given the reaction:

2 H2 + O2 = 2 H2O

2 * (2.01 ) g H2 ------------- 2 * ( 18.01 ) g H2O

mass H2 --------------------- 1.80 g H2O

mass H2 = 1.80 * 2 * 2.01 / 2* 18.01

mass H2 = 7.236 / 36.02

mass H2 = 0.2008 g

Hope that helps!

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We are given:

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So, mole fraction of $C_2H_4$ = 0.3

Mass percentage of $C_2H_2$ = 35 %

So, mole fraction of $C_2H_2$ = 0.35

Mass percentage of $C_2H_2O$ = 15 %

So, mole fraction of $C_2H_2O$ = 0.15

We know that:

Molar mass of $CH_4$ = 16 g/mol

Molar mass of $C_2H_4$ = 28 g/mol

Molar mass of $C_2H_2$ = 26 g/mol

Molar mass of $C_2H_2O$ = 48 g/mol

To calculate the average molecular mass of the mixture, we use the equation:

$\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}$

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$\chi_i$ = mole fractions of i-th species

$m_i$ = molar masses of i-th species

$n_i$ = number of observations

Putting values in above equation:

$\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}$

$\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75$

Hence, the correct answer is Option d.

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