The reaction between HNO3 and Ba(OH)2 is given by the equation below;
2HNO3 + Ba(OH)2 = Ba(NO3)2 + 2H2O
Moles of Barium hydroxide used;
= 0.200 × 0.039 l
= 0.0078 Moles
The mole ratio of HNO3 and Ba(OH)2 is 2: 1
Therefore; moles of nitric acid used will be;
= 0.0078 ×2 = 0.0156 moles
But; 0.0156 moles are equal to a volume of 0.10
The concentration of Nitric acid will be;
= (0.0156 × 1)/0.1
= 0.156 M
This is a incomplete question.The complete question is:
A chemist adds 180.0 ml of a 1.77 mol/L of sodium thiosulfate solution to a reaction flask. Calculate the mass in grams of sodium thiosulfate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
Answer: 50.4 g
Explanation:
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of sodium thiosulfate solution = 1.77 M
Volume of sodium thiosulfate solution = 180.0 mL = 0.1800 L
Putting values in equation 1, we get:
Mass of sodium thiosulfate = 
Thus 50.4 g of sodium thiosulfate the chemist has added to the flask.
Answer:
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Balanced equation:
<span>2 NO + 5 H2 ------> 2 NH3 + 2 H2O
</span>
<span>2 moles NO react with 5 moles H2 to produce 2 moles NH3
</span>
<span>Molar mass of NO = 30.00 g/mol </span>
<span>86.3g NO = 86.3/30.00 = 2.877 moles of NO </span>
<span>This will require: 2.877*5 / 2 = 7.192 moles of H2 </span>
<span>Molar mass of H2 = 2 g/mol </span>
<span>25.6g H2 = 25.6/2 = 12.7 mol H2. </span>
<span>You have excess H2 means the NO is limiting </span>
<span>From the balanced equation: </span>
<span>2 moles of NO will produce 2 moles of NH3 </span>
<span>2.877 moles of NO will produce 2.877 moles of NH3 </span>
<span>Molar mass NH3 = 17g/mol </span>
<span>Mass NH3 produced = 2.877 * 17 = 48.91g
Hence the yield is = 48.91 g ~ 49 g</span>
