Question

A mixer containing 64g of H2 and 64g of O2 is ignited so that water is formed as below:
2H2 + O2 -> 2H20
Identify the limiting reagent and calculate the amount of water formed and unreacted reactant.

Answer 1

$\huge{ \red{ \bigstar{ \purple{ \mathfrak{ \: \: answer}}}}}$

2H2+O2-------2H2o hence moles of h2 = 64/2 and moles of o2 is 64/32=2 SO it is clear that oxygen is limiting reagent so that 1mole of o2 gives two moles of water.

Hence 2*2*4 gives 72 gram of H2O and 56g of H2 so by this option 2 and 3 is correct

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Answer 2

$\huge\red{\underline{{\boxed{\textbf{QUESTION}}}}}$

A mixer containing 64g of H2 and 64g of O2 is ignited so that water is formed as below:2H2 + O2 -> 2H20Identify the limiting reagent and calculate the amount of water formed and unreacted reactant.

$\huge\red{\underline{{\boxed{\textbf{ANSWER}}}}}$

2H2+O2-------2H2o hence moles of h2 = 64/2 and moles of o2 is 64/32=2 SO it is clear that oxygen is limiting reagent so that 1mole of o2 gives two moles of water.. Hence 2*2*4 gives 72 gram of H2O and 56g of H2

H2 is limiting agent

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HARSH PRATAP SINGH