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2 years ago

What are the missing coefficients for the skeleton equation below? Cr(s)+Fe(NO3)2(aq)>Fe(s)+Cr(NO3)3(aq)

Chemistry

1 answer:

Stolb23 [73]2 years ago

7 0

Answer:

2 Cr(s) + 3 Fe(NO3)2(aq) = 3 Fe(s) + 2 Cr(NO3)3(aq)

Explanation:

I balanced this chemical equation so that both sides are equal, meaning that the mass of reactants is equal to mass of products.

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For the following reaction, 22.6 grams of nitrogen monoxide are allowed to react with 4.64 grams of hydrogen gas . nitrogen mono

antiseptic1488 [7]

Answer:

- 10.5 g of N₂

- Limiting reagent: NO

- 3.13 g of H₂ remains

Explanation:

First of all we state the reaction: 2NO(g) + 2H₂(g) → 2H₂O(l) + N₂(g)

We need to find out the limiting reactant and the excess reagent

Ratio in the reactants is 2:2. Let's convert the mass to moles:

22.6 g / 30 g/mol = 0.753 moles of NO

4.64 g / 2 g/mol = 2.32 moles of H₂

Certainly the limiting reagent is the NO and the excess reactant is the hydrogen:

- For 0.753 moles of NO, we need 0.753 moles of H₂ (we have 2.32 moles)

- For 2.32 moles of H₂, we need 2.32 moles of NO (and we don't have enough NO, because we only have 0.753 moles)

As the H₂ is the excess reagent, some moles still remains after the reaction is complete → 2.32 mol - 0.753 mol = 1.567 moles

We convert the moles to mass: 1.567 mol . 2g /1mol = 3.13 g of H₂ remains

As the NO is the limiting reagent, we can work with the equation:

We propose this rule of three: 2 moles of NO can produce 1 mol of N₂

Then, 0.753 moles of NO must produce (0.753 . 1) /2 = 0.376 moles of N₂

We convert the moles to mass 0.376 mol . 28 g / 1 mol = 10.5 g

3 0

3 years ago

What is the empirical formula

lord [1]

A formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.

5 0

3 years ago

Please help me get the answer to bothe a and b

neonofarm [45]

Answer:

If a metal and metal solution react, the more reactive metal will displace the less reactive metal from solution. If the metal in solution you start with is formed from a more reactive metal than the metal to be added, no reaction will occur.

3 0

1 year ago

A 0.753 g sample of a monoprotic acid is dissolved in water and titrated with 0.250 M NaOH. What is the molar mass of the acid i

Alexeev081 [22]

Answer:

$MM_{acid}=140.1g/mol$

Explanation:

Hello,

In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:

$n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}$

Thus, solving for the moles of the acid, we obtain:

$n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol$

Then, by using the mass of the acid, we compute its molar mass:

$MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol$

Regards.

7 0

2 years ago

In Part A, you found the amount of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Par

Finger [1]

First step is to balance the reaction equation. Hence we get P4 + 5 O2 => 2 P2O5

Second, we calculate the amounts we start with

P4: 112 g = 112 g/ 124 g/mol – 0.903 mol

O2: 112 g = 112 g / 32 g/mol = 3.5 mol

Lastly, we calculate the amount of P2O5 produced.

2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4 mol of P2O5.

This is 1.4 * (31*2 + 16*5) = 198.8 g

3 0

3 years ago