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3 years ago

HELPPP ... short response

Chemistry

1 answer:

Goshia [24]3 years ago

8 0

Answer:

Matter cannot be created or destroyed in chemical reactions.

Explanation:

Matter cannot be created or destroyed in chemical reactions. This is the law of conservation of mass. In every chemical reaction, the same mass of matter must end up in the products as started in the reactants. Balanced chemical equations show that mass is conserved in chemical reactions

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What is the percent yield of a reaction in which 51.5 g of tungsten(VI) oxide (WO3) reacts with excess hydrogen gas to produce m

Rus_ich [418]

Answer:

The percent yield of a reaction is 48.05%.

Explanation:

$WO_3+3H_2\rightarrow W+3H_2O$

Volume of water obtained from the reaction , V= 5.76 mL

Mass of water = m = Experimental yield of water

Density of water = d = 1.00 g/mL

$M=d\times V = 1.00 g/mL\times 5.76 mL=5.76 g$

Theoretical yield of water : T

Moles of tungsten(VI) oxide = $\frac{51.5 g}{232 g/mol}=0.2220 mol$

According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2220 moles of tungsten(VI) oxide will give:

$\frac{3}{1}\times 0.2220 mol=0.6660 mol$

Mass of 0.6660 moles of water:

0.666 mol × 18 g/mol = 11.988 g

Theoretical yield of water : T = 11.988 g

To calculate the percentage yield of reaction , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{m}{T}\times 100=\frac{5.76 g}{11.988 g}\times 100=48.05\%$

The percent yield of a reaction is 48.05%.

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Balance the equations by inserting coefficients as needed.

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Answer:

1. CaCO3 + 2HCl → CaCl2 + H2O + CO2

2. C6H12O2 + 8O2 → 6CO2 + 6H2O

Explanation:

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A sample of oxide iron weighing 2.40g was heated in a stream of hydrogen until it was completely converted to the metal. If the

wlad13 [49]

Answer:

Fe₂O₃

Explanation:

To solve this question we must find the moles of Iron in 1.68g. With the difference of the masses we can find the moles of oxygen. The formula will be obtained with the ratio of both amount of moles:

<em>Moles Fe:</em>

1.68g * (1mol / 56g) =0.03moles

<em>Moles O:</em>

2.40g-1.68g = 0.72g * (1mol/16g) = 0.045moles

The ratio O/Fe is:

0.045moles / 0.03moles = 1.5 moles. this ratio is obtained if the formula is:

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