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VMariaS [17]
3 years ago
12

2 H2 + O2 → 2 H2O How many grams of H2O are produced when 2.50 moles of oxygen are used?

Chemistry
1 answer:
vodomira [7]3 years ago
8 0

Answer:

90g

Explanation:

The equation for the reaction is given below:

2H2 + O2 → 2H2O

From the equation above,

1 mole of O2 produced 2 moles of H2O.

Therefore, 2.5 moles of O2 will produce = 2.5 x 2 = 5 moles of H2O.

Now let us convert 5 moles to H2O to grams to obtain the desired result. This is shown below:

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Number of mole of H2O = 5 moles

Mass of H2O = ?

Mass = number of mole x molar Mass

Mass of H2O = 5 x 18 = 90g

Therefore, 90g of H2O are produced

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What is the net ionic equation of the reaction of MgSO4 with Pb(NO3)2? Express you answer as a chemical equation including phase
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Answer : The net ionic equation will be,

Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The same number of ions present on reactant and product side which do not participate in a reactions.

The given balanced ionic equation will be,

MgSO_4(aq)+Pb(NO_3)_2(aq)\rightarrow Mg(NO_3)_2(aq)+PbSO_4(s)

The ionic equation in separated aqueous solution will be,

Mg^{2+}(aq)+SO_4^{2-}(aq)+Pb^{2+}(aq)+2NO^{3-}(aq)\rightarrow PbSO_4(s)+Mg^{2+}(aq)+2NO^{3-}(aq)

In this equation, Mg^{2+}\text{ and }NO_3^- are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)

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Heating 2.40 g of the oxide of metal X (molar mass of X = 55.9 g/mol) in carbon monoxide (CO) yields the pure metal and carbon d
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Answer:The molecular formula of the oxide of metal be X_2O_3. The balanced equation for the reaction is given by:

X_2O_3+3CO\rightarrow 3CO_2+2X

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Let the molecular formula of the oxide of metal be X_2O_y

X_2O_y+yCO\rightarrrow yCO_2+2X

Mass of metal product = 1.68 g

Moles of metal X =\frac{1.68 g}{55.9 g/mol}=0.03005 mol

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Then 0.03005 moles of metal X will be produced by:

\frac{1}{2}\times 0.03005 mol=0.01502 mol of metal oxide

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