Butane is C₄H₁₀.

$C_4H_{10} + O_2 \to CO_2 + H_2O \\ \\
\hbox{balance carbon and hydrogen on the right-hand side:} \\
C_4 H_{10} + O_2 \to 4 \ CO_2 + 5 \ H_2O \\ \\
\hbox{balance oxygen on the left-hand side:} \\
C_4 H_{10} + \frac{13}{2} \ O_2 \to 4 \ CO_2 + 5 \ H_2O \\ \\
\hbox{multiply by 2 to get rid of the fraction:} \\
2 \ C_4H_{10} + 13 \ O_2 \to 8 \ CO_2 + 10 \ H_2O$

The balanced equation is 2 C₄H₁₀ + 13 O₂ <span>→</span> 8 CO₂ + 10 H₂O.

3 0

3 years ago

Find the mass, in grams, of 3.758 mol CH4. ( show work )

vladimir2022 [97]

Answer:

The mass of CH4 is 60, 29 grams.

Explanation:

We use the weight of the atoms C and H for calculate the molar mass:

Weight of CH4= weight C+ 4 x weight H= 12,01 g/mol +4 x 1,008g/mol=

Weight of CH4 =16, 042 g/mol

1molCH4-----16, 042grams

3,758 mol CH4--X= (3,758 mol CH4 x 16, 042 grams)/1 mol CH4=60,285836 grams

5 0

3 years ago

When 1000 C of charge is passed through CuSO4 solution, x g of copper is deposited. How much charge should be passed through the

Stels [109]

Number of charge = 5018 C

<h3>Further explanation</h3>

Given

1000 C of charge for x grams of copper

Required

Number of charge

Solution

Faraday's Law :

$\tt W=\dfrac{e.i.t}{96500}\\\\W=\dfrac{e.Q}{96500}$

For 1000 C, W = x grams

$\tt x=\dfrac{1000.e}{96500}=0.0104e$

For 5x grams :

$\tt 5x=\dfrac{e.Q}{96500}\\\\5\times 0.0104e=\dfrac{e.Q}{96500}\\\\Q=\dfrac{96500\times 5\times 0.0104e}{e}=5018~C$

6 0

3 years ago