Question

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

Answer 1

Answer : The correct option is, (C) 0.200 g

Explanation : Given,

Mass of water = 1.80 g

Molar mass of water = 18 g/mole

Molar mass of $H_2$ = 2 g/mole

First we have to calculate the moles of water.

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{1.80g}{18g/mole}=0.1mole$

Now we have to calculate the moles of hydrogen gas.

The given balanced reaction is,

$2H_2+O_2\rightarrow 2H_2O$

From the balanced reaction we conclude that

As, 2 mole of water obtained from 2 mole of $H_2$ gas

So, 0.1 mole of water obtained from 0.1 mole of $H_2$ gas

Now we have to calculate the mass of hydrogen gas.

$\text{Mass of }H_2=\text{Moles of }H_2\times \text{Molar mass of }H_2$

$\text{Mass of }H_2=(0.1mole)\times (2g/mole)=0.2g$

Therefore, the mass of hydrogen gas required will be, 0.2 grams

Answer 2

Answer is c

Work out the moles of 2H2O: 1.80/ (16+1+1) = 0.1
The moles of 2H2 are the same (0.1)
The Mr of H2 is 2

So the mass needed to produce 1.80g of water is
0.1 x 2 = 0.2