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3 years ago

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The chemical equation below shows the process of forming water. Balance the equation by calculating the coefficients. H2 + O2 H2

O A. 2, 1, 2 B. 1, 1, 1 C. 1, 1, 2 D. 1, 2, 2

1 answer:

Mrrafil [7]3 years ago

5 0

Answer: A. 212

Explanation:

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The fact that HBO2, a reactive compound, was produced rather than the relatively inert B2O3 was a factor in the discontinuation

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Answer:

1027.62 g

Explanation:

For $B_2H_6$ :-

Mass of $B_2H_6$ = 296.1 g

Molar mass of $B_2H_6$ = 27.66 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{296.1\ g}{27.66\ g/mol}$

$Moles\ of\ B_2H_6= 10.705\ mol$

From the balanced reaction:-

$B_2H_6(g) + 3 O2_{(l)}\rightarrow 2 HBO_2_{(g)}+ 2 H_2O_{(l)}$

1 mole of $B_2H_6$ react with 3 moles of oxygen

Thus,

10.705 mole of $B_2H_6$ react with 3*10.705 moles of oxygen

Moles of oxygen = 32.115 moles

Molar mass of oxygen gas = 31.998 g/mol

<u>Mass = Moles * Molar mass = 32.115 * 31.998 g = 1027.62 g</u>

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3 years ago

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Usimov [2.4K]

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3 years ago

Read 2 more answers

If 7.0 mol sample of a gas has a volume of 12.2 L, what would the volume be if the amount of gas was increased to 16.8 mol

Leviafan [203]

Answer:

$V_{2} = 29.28\,L$

Explanation:

Let assume that gas behaves ideally and experiments an isobaric and isothermal processes. The following relationship is applied to determined the final volume:

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$

$V_{2} = V_{1} \cdot \left(\frac{n_{2}}{n_{1}} \right)$

$V_{2} = (12.2\,L)\cdot \left(\frac{16.8\,moles}{7\,moles} \right)$

$V_{2} = 29.28\,L$

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