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2 years ago

How many grams of hydrogen are needed to produce 1.80 g of water according to this equation? 2H2 + O2 → 2H2O

Chemistry

1 answer:

Arisa [49]2 years ago

7 0

Hey there!:

Molar mass:

H2 = 2.01 g/mol ; H2O = 18.01

Given the reaction:

2 H2 + O2 = 2 H2O

2 * (2.01 ) g H2 ------------- 2 * ( 18.01 ) g H2O

mass H2 --------------------- 1.80 g H2O

mass H2 = 1.80 * 2 * 2.01 / 2* 18.01

mass H2 = 7.236 / 36.02

mass H2 = 0.2008 g

Hope that helps!

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Based on the information above, which statement provides evidence that a chemical reaction has occurred?

Vesna [10]

Answer: Atoms that make up reactants rearrange into new products

Explanation:

A chemical reaction is the process of breaking down the chemical bonds of the reacting substances which are commonly called as reactants and making new bonds to form new substances commonly called as products.

The products will contain the same atoms as there are in the reactants but the bonds will rearrange to form different substances.

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2 years ago

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

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Answer: The concentration of $NH_3$ required will be 0.285 M.

Explanation:

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

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