Methane burns in air according to this equation:

Question

An equation.Short description, An equation.,Long description,
Upper C Upper H 4 left-parenthesis g right parenthesis plus 2 Upper O 2 left-parenthesis g right parenthesis yields Upper C Upper O 2 left-parenthesis g right parenthesis plus 2 Upper H 2 Upper O left-parenthesis l right parenthesis



The table shows the molar masses of substances involved in the reaction.



Molar Masses
,begin bold,Compound,end bold, ,begin bold,Molar Mass,end bold, (grams per mole)
Upper C Upper H Subscript 4 16.0
Upper C Upper O Subscript 2 44.0
Upper H Subscript 2 Upper O 18.0
Upper O Subscript 2 32.0

Question
How many grams of water will be produced if 64.0 grams of Upper C Upper H Subscript 4 burns in excess Upper O Subscript 2?

Answer options with 4 options
A.
8.0 grams

B.
36.0 grams

C.
72.0 grams

D.
144.0 grams

Answer 1

To determine how many grams of water (\(H_2O\)) will be produced when 64.0 grams of methane (\(CH_4\)) burns, we can follow these steps:

1. Write the balanced chemical equation: \[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \]

2. Calculate the moles of \(CH_4\) in 64.0 grams: \[ \text{Molar mass of } CH_4 = 16.0 \text{ g/mol} \] \[ \text{Moles of } CH_4 = \frac{64.0 \text{ g}}{16.0 \text{ g/mol}} = 4.0 \text{ moles} \]

3. Using the stoichiometry of the reaction, determine the moles of \(H_2O\) produced: According to the balanced equation, 1 mole of \(CH_4\) produces 2 moles of \(H_2O\). Therefore: \[ \text{Moles of } H_2O = 4.0 \text{ moles of } CH_4 \times \frac{2 \text{ moles of } H_2O}{1 \text{ mole of } CH_4} = 8.0 \text{ moles of } H_2O \]

4. Calculate the mass of \(H_2O\) produced: \[ \text{Molar mass of } H_2O = 18.0 \text{ g/mol} \] \[ \text{Mass of } H_2O = 8.0 \text{ moles} \times 18.0 \text{ g/mol} = 144.0 \text{ grams} \]

Thus, the answer is: D. 144.0 grams.