What is the number of grams in .5 mol of H2S04

The balanced equation for combustion in an acetylene torch is shown below: 2C2H2 + 5O2 → 4CO2 + 2H2O The acetylene tank contains

andriy [413]

Answer: 67.2 moles

Explanation: $2C_2H5+5O_2\rightarrow 4CO_2+2H_2O$

According to the given balanced equation, 2 moles of acetylene $C_2H_2$ combine with 5 moles of oxygen $O_2$ to produce 4 moles of carbon dioxide $CO_2$ .

Thus if 2 moles of acetylene $C_2H_2$ combine with = 5 moles of oxygen $O_2$

35 moles of acetylene $C_2H_2$ combine with= $\frac{5}{2}\times {35}=87.5$ moles of oxygen $O_2$

But as only 84 moles of oxygen are available, acetylene is not a limiting reagent.

5 moles of oxygen $O_2$ reacts with = 2 moles of acetylene $C_2H_2$

84 moles of oxygen $O_2$ reacts with= $\frac{2}{5}\times {84}=33.6$ moles of acetylene $C_2H_2$

Thus Oxygen is the limiting reagent as it limits the formation of products. Acetylene is excess reagent as it is present in excess.

2 moles of acetylene $C_2H_2$ produce= 4 moles of carbon dioxide $CO_2$ .

33.6 moles of acetylene $C_2H_2$ produce= $\frac{4}{2}\times {33.6}=67.2$ moles of of carbon dioxide $CO_2$ .

8 0

3 years ago

Read 2 more answers

CaBr + KOH – Ca(OH), + KBr (balance first) What mass, in grams, of

neonofarm [45]

Answer:

129.73 g of CaBr₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CaBr₂ + 2KOH –> Ca(OH)₂ + 2KBr

Next, we shall determine the mass of CaBr₂ that reacted and the mass of Ca(OH)₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of CaBr₂ = 40 + (80×2)

= 40 + 160

= 200 g/mol

Mass of CaBr₂ from the balanced equation = 1 × 200 = 200 g

Molar mass of Ca(OH)₂ = 40 + 2(16 + 1)

= 40 + 2(17)

= 40 + 34

= 74 g/mol

Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g

SUMMARY :

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Finally, we shall determine the mass of CaBr₂ that react when 48 g of Ca(OH)₂ were produced. This can be obtained as follow:

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Therefore, Xg of CaBr₂ will react to produce 48 g of Ca(OH)₂ i.e

Xg of CaBr₂ = (200 × 48)/74

Xg of CaBr₂ = 129.73 g

Thus, 129.73 g of CaBr₂ were consumed.

6 0

3 years ago

CALCULATOR

suter [353]

Answer: The possible molecular formula will be $CO_2$

Explanation:

Mass of C= 27.3 g

Mass of O = 72.7 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{27.3g}{12g/mole}=2.275moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{72.7g}{16g/mole}=4.544moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{2.275}{2.275}=1$

For O = $\frac{4.544}{2.275}=2$

The ratio of C : O = 1: 2

Hence the empirical formula is $CO_2$

The possible molecular formula will be= $n\times CO_2$

5 0

2 years ago

3. For each equation, state the reaction type and then balance the equation.

Natali5045456 [20]

2 KCIO3 -> 2 KCI + 3 O2

4 0

2 years ago

Which statement(s) correctly compare the masses of protons, neutrons, and electrons? Select two options.

Ludmilka [50]

Answer:

6 Electrons r smaller than a proton or a neutron

5 0

2 years ago