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Percentage Composition

90 g mixture ...

Question

$90$ g mixture of $H_{2}$ and $O_{2}$ is taken in stoichiometric ratio and gives $H_{2} O$ with $50 %$ yield. The produced mass of $H_{2} O$ (in gm) is:

45

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36

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20

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90

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Solution

The correct option is A $45$ g
Since 90g mixture gives 50 percent by mass yield of product i.e. $H_{2} O$ .
The mass of $H_{2} O$ produced would directly become 45g.

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Similar questions

(O_{2})

0.20 g

(H_{2})

(H_{2} O)

H_{2} = 2 g/mol, O_{2} = 32 g/mol

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} H_{2} (g) + \frac{1}{2} O_{2} (g) = H_{2} O (g); Δ H = - 241.8 k J C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) = 2 C O_{2} (g) + H_{2} O (g); Δ H = - 1300 k J \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

C_{2} H_{2}

H_{2}

1.15 g

N a

0.9 g

H_{2} O

80 %

H_{2}

N_{2}

H_{2}

N H_{3}

N H_{3}

[C a (H C O_{3})_{2}]

C_{4} H_{10} + O_{2} \to C O_{2} + H_{2} O C a C O_{3} + C O_{2} + H_{2} O \to C a (H C O_{3})_{2}

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