Question

If 2 grams of H2 is reacted with excess O2, how many grams of H2O would be produced ?

Answer 1

18 g `H_2O`

Explanation:

Solution Map:
grams `H_2->`mol`O_2->`mol`H_2O->` grams`H_2O`
`H_2`=2.016 g/mol
`H_2O`=18.015 g/mol

`2cancel(gH_2)*(cancel(molH_2))/(2.016cancel(gH_2))*(cancel(molO_2))/(cancel(molH_2))*(cancel(molH_2O))/(cancel(molO_2))*(18.015gH_2O)/(cancel(molH_2O))`

Answer 2

The answer is 18 grams.

Explanation:

Let's start by writing the equation of reaction:

`H_"2"+O_"2"rarrH_"2"O`

The equation is not balanced, so lets balance it:

First balance the number of H atoms on both sides and then O atoms and we need atomic and molecular weights:

- Atomic weight of H = 1 gram
- Atomic weight of O = 16 grams

`2H_"2"+O_"2"rarr2H_"2"O`

So,
`2H_"2" = 4` grams
`O_"2" = 32` grams
`2H_"2"O = 36` grams

In the question we are given 2 grams of `H_"2"` and we have to calculate the weight of `H_"2"O`:

By the equation, we have:

`4` grams of `H_"2"` gives `36` grams of `H_"2"O`
or, `1` gram of `H_"2"` gives `36/4` grams of `H_"2"O`
or, `2` gram of `H_"2"` gives `(36/4)*2` grams of `H_"2"O`
Therefore, `2` grams of `H_"2"` gives `18` grams of `H_"2"O`.

Hope This Was Helpful....