A reaction requires two atoms of $P$ for five atoms of $O$ . The amount in grams of $O_{2}$ required for $3.10$ g $P$ is:

Open in App

Solution

According to given data,
$5$ atom $O$ $\equiv$ $2$ atom $P$
$5 \times 16$ $2 \times 31$
$?$ $3.10$
So, oxygen required is $= \frac{5 \times 16 \times 3.1}{2 \times 31} = 4$ g.

Suggest Corrections

Similar questions

Q. A reaction required three atoms of Mg for two atoms of N. How many grams of N are required for a 3.6 g of Mg?

Q. The energy required for the ionisation of 0.02 gram atom of magnesium is

x

kJ. The amount of energy required to ionise 1 atom of magnesium is:

Q. For the given disproportionation reaction:

H_{3} P O_{2} \to P H_{3} + H_{3} P O_{3}

The equivalent mass of

H_{3} P O_{2}

in grams is :
(Given :

Atomic mass of P = 31 u

)

$O_{2}$ undergoes photochemical dissociation into one normal and one excited atoms. The excited atom has 1.967 eV more energy than normal. The dissociation of $O_{2}$ into two normal atoms requires 498 KJ mole-1. Then the maximum wavelength required for photochemical dissociation of $O_{2}$ is

Q. What is the minimum quantity (in grams) of methyl iodide required for preparing

1

mole of ethane by Wurtz reaction?

[Atomic weight of iodine

= 127

]

Join BYJU'S Learning Program

A reaction requires two atoms of P for five atoms of O. The amount in grams of O2 required for 3.10 g P is:

According to given data,5 atom O ≡ 2 atom P5×16 2×31? 3.10So, oxygen required is =5×16×3.12×31=4 g.

A reaction requires two atoms of $P$ for five atoms of $O$ . The amount in grams of $O_{2}$ required for $3.10$ g $P$ is:

According to given data,
$5$ atom $O$ $\equiv$ $2$ atom $P$
$5 \times 16$ $2 \times 31$
$?$ $3.10$
So, oxygen required is $= \frac{5 \times 16 \times 3.1}{2 \times 31} = 4$ g.